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 

#ifndef ONLINE_JUDGE
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#else
#include <bits/stdc++.h>
#endif
using namespace std;

#define ms(s, n) memset(s, n, sizeof(s))
#define FOR(i, a, b) for(int i = a; i < b; i++)
#define FORd(i, a, b) for(int i = a - 1; i >= b; i--)
#define sz(a) int((a).size())
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<string> vs;
typedef vector<pi> vii;
typedef vector<pl> vll;
ll gcd(ll a, ll b) {ll r; while (b != 0) {r = a % b; a = b; b = r;} return a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}

const int maxn = 1000010;
const int mod = 1000000007;
int n, m;
char p[maxn];
int nlen;
int y[maxn];
ll hash[maxn];
ll po[maxn];

bool check(int i) {
	return (hash[i - 1] * po[nlen - i] - hash[nlen - 1] + hash[nlen - i - 1] + mod) % mod == 0;
}

void solve() {
	scanf("%d%d%s", &n, &m, p);
	nlen = strlen(p);
	po[0] = 1;
	FOR(i, 0, nlen) {
		if (i) hash[i] = (hash[i - 1] + p[i] * po[i]) % mod; else hash[i] = p[i];
		po[i + 1] = po[i] * 257 % mod;
	}
	int cnt = 0;
	FOR(i, 0, m) {
		scanf("%d", y + i);
		if (i && y[i] - y[i - 1] < nlen && !check(nlen - y[i] + y[i - 1])) {printf("0"); return;}
		if (i && y[i] - y[i - 1] > nlen) {cnt += y[i] - y[i - 1] - nlen;}
	}
	if (m) cnt += n + y[0] - (y[m - 1] + nlen); else cnt = n;
	ll ans = 1;
	FOR(i, 0, cnt) ans = ans * 26 % mod;
	printf("%I64d", ans);
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout);
#endif
	solve();
	return 0;
}