Tavas is a cheerleader in the new sports competition named "Pashmaks".

<image>

This competition consists of two part: swimming and then running. People will immediately start running R meters after they finished swimming exactly S meters. A winner is a such person that nobody else finishes running before him/her (there may be more than one winner).

Before the match starts, Tavas knows that there are n competitors registered for the match. Also, he knows that i-th person's swimming speed is si meters per second and his/her running speed is ri meters per second. Unfortunately, he doesn't know the values of R and S, but he knows that they are real numbers greater than 0.

As a cheerleader, Tavas wants to know who to cheer up. So, he wants to know all people that might win. We consider a competitor might win if and only if there are some values of R and S such that with these values, (s)he will be a winner.

Tavas isn't really familiar with programming, so he asked you to help him.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 2 × 105).

The next n lines contain the details of competitors. i-th line contains two integers si and ri (1 ≤ si, ri ≤ 104).

Output

In the first and the only line of output, print a sequence of numbers of possible winners in increasing order.

Examples

Input

3
1 3
2 2
3 1


Output

1 2 3 


Input

3
1 2
1 1
2 1


Output

1 3 
5
10 50
10 50
10 50
10 50
10 50
 2
10000 10000
10000 10000
 3
10000 10000
10000 10000
10000 10000
 3
1000 3000
1500 1500
3000 1000
 43
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 1
1 1
 2
1 1
1 1
 44
78 52
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 3
1000 3000
1499 1499
3000 1000
 5
50 50
49 50
50 49
49 49
50 1
 18
82 38
33 69
33 69
33 69
33 69
33 69
82 38
33 69
33 69
33 69
82 38
33 69
33 69
82 38
82 38
33 69
33 69
33 69
 1
10000 9999
 3
1000 3000
1501 1501
3000 1000
 2
1 2
2 1

1 2 3 4 5
 1 2
 1 2 3
 1 2 3
 1 3 16 24 29 31 32 37 41 42
 1
 1 2
 1
 1 3
 1
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
 1
 1 2 3
 1 2 

#define _CRT_SECURE_NO_WARNINGS
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <algorithm>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>

#define oo 1e9
#define pi 3.1415926536
#define all(x) x.begin(),x.end()
#define sorta(x) sort(all(x))
#define sortam(x,comp) sort(all(x),comp)
#define sortd(x) sort(x.rbegin(),x.rend())
#define sf(x) scanf("%d", &x);
#define sf2(x, y) scanf("%d %d", &x, &y);
#define sf3(x, y, z) scanf("%d %d %d", &x, &y, &z);
#define sfll(x) scanf("%I64d", &x);
#define sfll2(x, y) scanf("%I64d %I64d", &x, &y);
#define sfll3(x, y, z) scanf("%I64d %I64d %I64d", &x, &y, &z);
#define sfd(x) scanf("%f", &x);
#define pr(x) printf("%d ", x);

typedef long long ll;
using namespace std;

ll mod = 1e9 + 7;
ll pw(ll b, ll p) {
    if(p < 0) return 0;
    if(!p) return 1;
    if(p == 1) return b;
    if(p % 2) return (b*pw((b*b) % mod, (p - 1) / 2) % mod) % mod;
    return pw((b*b) % mod, p / 2) % mod;
}

vector<int>Z(string s) {
    int l = 0, r = 0, n = s.size();
    vector<int>z(n, 0);
    for(int i = 1; i < n; i++) {
        if(i>r) {
            l = r = i;
            while(r < n && s[r - l] == s[r]) r++;
            z[i] = r - l;
            r--;
        }
        else {
            int k = i - l;
            if(z[k] < r - i + 1) z[i] = z[k];
            else {
                l = i;
                while(r < n && s[r - l] == s[r]) r++;
                z[i] = r - l;
                r--;
            }
        }
    }
    return z;
}

int main() {
    int n, m;
    while(cin >> n >> m) {
        string s;
        cin >> s;
        vector<ll>a(m);
        ll p = 0;
        for(int i = 0; i < m; i++) {
            sfll(a[i]);
            if(i) p += max(0LL, a[i] - (a[i - 1] + s.size()));
            else p += a[i] - 1;
            if(i == m - 1) p += n - (a[i] + s.size()) + 1;
        }

        if(!m) {
            cout << pw(26, n) << endl;
            continue;
        }

        vector<int>z = Z(s);
        for(int i = 1; i < m; i++) {
            if(a[i] <= a[i - 1] + s.size() && z[a[i] - a[i - 1]] != s.size() - (a[i] - a[i - 1])) {
                p = -1;
                break;
            }
        }
        cout << pw(26, p) % mod << endl;
    }
    return 0;
}
