Problem: 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

Solution (in JAVA):

Solution:
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.NoSuchElementException;

public class Main {
	static final int MOD = (int)1e9 + 7;
	int N,M;
	char[] ch;
	int[] r,l,newR,sum;
	long[][] dp;

	public void solve() {
		N = nextInt();
		M = nextInt();
		ch = next().toCharArray();

		l = new int[M];
		r = new int[M];
		for(int i = 0;i < M;i++){
			l[i] = nextInt() - 1;
			r[i] = nextInt() - 1;
		}

		newR = new int[N];
		for(int i = 0;i < N;i++){
			newR[i] = i;
		}
		for(int i = 0;i < M;i++){
			newR[l[i]] = Math.max(newR[l[i]],r[i]);
		}

		for(int i = 1;i < N;i++){
			newR[i] = Math.max(newR[i], newR[i-1]);
		}

		sum = new int[N];
		for(int i = 0;i < N;i++){
			sum[i] = ch[i] - '0';
		}
		for(int i = 0;i < N - 1;i++){
			sum[i + 1] += sum[i];
		}

		dp = new long[N + 1][N + 1];
		dp[0][0] = 1L;

		//i = 今見ているSの位置
		for(int i = 0;i < N;i++){

			//j = これまで配置した1の数
			for(int j = 0;j <= i;j++){

				if(dp[i][j] == 0)continue;

				/*
				 * j <= sum[newR[i]] jがnewR[i]]の位置までに存在する1の数以下(これを超えていたら存在する1の数より多いのでありえない)
				 * sum[newR[i]] - j <= newR[i] - i 残りの配置しなければならない1の数 <= newR[i]の位置まで配置できる0or1の数(これも超えていたら1が余分に存在している)
				 *
				 * 以上の条件を満たした時、dp[i + 1][j]に遷移する(iの位置に0を配置して次の状態に移る)
				 */
				if(j <= sum[newR[i]] && sum[newR[i]] - j <= newR[i] - i){
					dp[i + 1][j] += dp[i][j]%MOD;
					dp[i + 1][j] %= MOD;
				}

				/*
				 * j <= sum[newR[i]] - 1 iの位置に1を配置する分をsum[newR[i]]から引いてもちゃんとj以上になる(これがj未満になると矛盾していることになる)
				 * sum[newR[i]] - j - 1 <= newR[i] - i 今iの位置に配置する分(-1)とこれまで配置した分(j)をsum[newR[i]]から引いた残りの数(これはnewR[i]の位置までに配置しなければならない1の数)が
				 * 残りの配置できる位置の数以下(これが上回っていると配置することが出来ない1が出てきてしまう)
				 */
				if(j  + 1<= sum[newR[i]] && sum[newR[i]] - j - 1 <= newR[i] - i){
					dp[i + 1][j + 1] += dp[i][j]%MOD;
					dp[i + 1][j + 1] %= MOD;
				}
			}
		}

		out.println(dp[N][sum[N-1]]);
	}

	public static void main(String[] args) {
		out.flush();
		new Main().solve();
		out.close();
	}

	/* Input */
	private static final InputStream in = System.in;
	private static final PrintWriter out = new PrintWriter(System.out);
	private final byte[] buffer = new byte[2048];
	private int p = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (p < buflen)
			return true;
		p = 0;
		try {
			buflen = in.read(buffer);
		} catch (IOException e) {
			e.printStackTrace();
		}
		if (buflen <= 0)
			return false;
		return true;
	}

	public boolean hasNext() {
		while (hasNextByte() && !isPrint(buffer[p])) {
			p++;
		}
		return hasNextByte();
	}

	private boolean isPrint(int ch) {
		if (ch >= '!' && ch <= '~')
			return true;
		return false;
	}

	private int nextByte() {
		if (!hasNextByte())
			return -1;
		return buffer[p++];
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = -1;
		while (isPrint((b = nextByte()))) {
			sb.appendCodePoint(b);
		}
		return sb.toString();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}
}