Codeforces 395 D.Pair of Numbers

D. Pair of Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l, r (1 ≤ l ≤ r ≤ n), such that the following conditions hold:

1. there is integer j (l ≤ j ≤ r), such that all integers al, al + 1, ..., ar are divisible by aj;
2. value r - l takes the maximum value among all pairs for which condition 1 is true;

Help Simon, find the required pair of numbers (l, r). If there are multiple required pairs find all of them.

Input

The first line contains integer n (1 ≤ n ≤ 3·105).

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 106).

Output

Print two integers in the first line — the number of required pairs and the maximum value of r - l. On the following line print all l values from optimal pairs in increasing order.

Examples

input

5
4 6 9 3 6

output

1 3
2 

input

5
1 3 5 7 9

output

1 4
1 

input

5
2 3 5 7 11

output

5 0
1 2 3 4 5 

Note

In the first sample the pair of numbers is right, as numbers 6, 9, 3 are divisible by 3.

In the second sample all numbers are divisible by number 1.

In the third sample all numbers are prime, so conditions 1 and 2 are true only for pairs of numbers (1, 1), (2, 2), (3, 3), (4, 4), (5, 5).

题目大意：给出长度为n的序列a[i]，要求找到所有满足下列两个条件的子序列a[l],a[l+1],…,a[r]的个数： 
1.存在l<=j<=r，使得a[j]是a[l],a[l+1],…,a[r]的最大公因数 
2.在所有满足1的子序列中取r-l最长的.

分析：一个序列满足要求当且仅当min = gcd,关键的问题就是如何快速地求出min和gcd,最好是O(1),直接ST表就可以了.接下来二分枚举区间.一般这种求有多少个区间的都是先固定一个左端点然后二分右端点，但是这道题中min和gcd的单调性相同，不能直接这样二分，题目要求最大长度，那么直接二分长度，再枚举起点判断即可.

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 300010;
int n, f[maxn][22], g[maxn][22], ans, anss[300010], tot, cnt, t[300010], anscnt;

int gcd(int a, int b)
{
    if (!b)
        return a;
    return gcd(b, a % b);
}

void init()
{
    for (int j = 1; j <= 20; j++)
        for (int i = 1; i + (1 << j) - 1 <= n; i++)
        {
            f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
            g[i][j] = gcd(g[i][j - 1], g[i + (1 << (j - 1))][j - 1]);
        }
}

int geta(int l, int r)
{
    int k = (int)((log(r - l + 1)) / log(2.0));
    return min(f[l][k], f[r - (1 << k) + 1][k]);
}

int getb(int l, int r)
{
    int k = (int)((log(r - l + 1)) / log(2.0));
    return gcd(g[l][k], g[r - (1 << k) + 1][k]);
}

bool solve(int x)
{
    cnt = 0;
    int tott = 0;
    memset(t, 0, sizeof(t));
    for (int i = 1; i + x <= n; i++)
        if (geta(i, i + x) == getb(i, i + x))
        {
            cnt++;
            t[++tott] = i;
        }
    if (cnt)
    {
        if (x > ans)
        {
            memcpy(anss, t, sizeof(t));
            tot = tott;
            anscnt = cnt;
        }
        return true;
    }
    return false;
}

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &f[i][0]);
        g[i][0] = f[i][0];
    }
    init();
    int l = 0, r = n;
    while (l <= r)
    {
        int mid = (l + r) >> 1;
        if (solve(mid))
        {
            l = mid + 1;
            ans = mid;
        }
        else
            r = mid - 1;
    }
    if (ans != 0)
    {
        printf("%d %d\n", anscnt, ans);
        for (int i = 1; i <= tot; i++)
            printf("%d ", anss[i]);
    }
    else
    {
        printf("%d %d\n", n, ans);
        for (int i = 1; i <= n; i++)
            printf("%d ", i);
    }

    return 0;
}