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>
using namespace std;

#define all(s) s.begin(), s.end()
#define pb push_back
#define ii pair<int, int>
#define x first
#define y second
#define bit(x, y) ((x >> y) & 1)

const int MOD = 1000000007;

int dp[3005][3005];

int main() {
    ios::sync_with_stdio(false); cin.tie(0); 
    cout.tie(0);
    int n, m;
    cin >> n >> m;
    string s;
    cin >> s;
    s = " " + s;
    string t = s;
    for (int i = 1; i <= m; i++) {
        int u, v;
        cin >> u >> v;
        sort(s.begin() + u, s.begin() + v + 1);
        sort(t.begin() + u, t.begin() + v + 1, [](char x, char y) {
            return x > y;
        });
    }
    vector<int> cntmx(n + 1);
    vector<int> cntmn(n + 1);
    for (int i = 1; i <= n; i++) {
        cntmx[i] = cntmx[i - 1] + (s[i] == '0');
        cntmn[i] = cntmn[i - 1] + (t[i] == '0');
    }

    dp[0][0] = 1;
    for (int i = 1; i <= n; i++) {
        if (cntmn[i] == 0) dp[i][0] = dp[i - 1][0];
        for (int j = max(cntmn[i], 1); j <= cntmx[i]; j++) {
            dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j]) % MOD;
        }
    }

    int ans = 0;
    for (int i = 0; i <= 3000; i++) {
        ans = (ans + dp[n][i]) % MOD;
    }

    cout << ans;

    return 0;
}