There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143
#include <bits/stdc++.h>
#define Task        "test"
#define pb          emplace_back
#define fi          first
#define se          second
#define mp          make_pair
#define ll          long long
//#define int         int64_t

using namespace std;

typedef pair<ll, ll> pii;
const int N = (int)3e3 + 2;
const int mod = (int)1e9 + 7;

int n, m, l, r, sum, f[N][N], cnt[N], jump[N];
string s;

void add(int& x, int y) {if((x += y) >= mod) x -= mod;}

int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    if(fopen(Task".inp", "r")) {
        freopen(Task".inp", "r", stdin);
        freopen(Task".out", "w", stdout);
    }
    cin >> n >> m >> s; s = ' ' + s;
    for(int i = 1; i <= m; ++i) {
        cin >> l >> r;
        jump[l] = max(jump[l], r);
    }
    for(int i = 1; i <= n + 1; ++i) {
        jump[i] = max({jump[i], jump[i - 1], i});
        cnt[i] = cnt[i - 1] + (s[i] == '1');
    }
    f[1][cnt[jump[1]]] = 1;
    for(int i = 1; i <= n + 1; ++i) {
        l = jump[i], r = jump[i + 1];
        sum = cnt[r] - cnt[l];
        for(int j = 0; j <= cnt[n]; ++j) {
            if(f[i][j] == 0) continue;
            if(j) add(f[i + 1][j - 1 + sum], f[i][j]);
            if(jump[i] - i + 1 - j > 0) add(f[i + 1][j + sum], f[i][j]);
        }
    }
    cout << f[n + 1][0];
}
