Problem: You are given an integer sequence of length N: A_1,A_2,...,A_N. Let us perform Q operations in order. The i-th operation is described by two integers X_i and Y_i. In this operation, we will choose one of the following two actions and perform it:

* Swap the values of A_{X_i} and A_{Y_i}
* Do nothing



There are 2^Q ways to perform these operations. Find the sum of the inversion numbers of the final sequences for all of these ways to perform operations, modulo 10^9+7.

Here, the inversion number of a sequence P_1,P_2,...,P_M is the number of pairs of integers (i,j) such that 1\leq i < j\leq M and P_i > P_j.

Constraints

* 1 \leq N \leq 3000
* 0 \leq Q \leq 3000
* 0 \leq A_i \leq 10^9(1\leq i\leq N)
* 1 \leq X_i,Y_i \leq N(1\leq i\leq Q)
* X_i\neq Y_i(1\leq i\leq Q)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
A_1
:
A_N
X_1 Y_1
:
X_Q Y_Q


Output

Print the sum of the inversion numbers of the final sequences, modulo 10^9+7.

Examples

Input

3 2
1
2
3
1 2
1 3


Output

6


Input

5 3
3
2
3
1
4
1 5
2 3
4 2


Output

36


Input

9 5
3
1
4
1
5
9
2
6
5
3 5
8 9
7 9
3 2
3 8


Output

425

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<functional>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
#define MOD 1000000007
#define f(i,n) for(int i=0;i<int(n);i++)
#define N 200000
long long a[3000][3000];





int main() {
	int n, q, x, y;
	long long z;
	int b[3000];
	scanf("%d %d", &n,&q);
	f(i, n)scanf("%d", &b[i]);
	f(i, n) {
		f(j, n)a[i][j] = 0;
	}
	f(i, n) {
		f(j, i) {
			if (b[j] < b[i]) {
				a[j][i] = 1;
			}
			else if (b[j] > b[i]) {
				a[i][j] = 1;
			}
		}
	}
	f(qq, q) {
		scanf("%d %d", &x, &y);
		x--;
		y--;
		f(i, n) {
			if ((i != x) && (i != y)) {
				z = a[i][x];
				z += a[i][y];
				z *= 500000004;
				z %= MOD;
				a[i][x] = z;
				a[i][y] = z;
			}
		}
		f(i, n) {
			if ((i != x) && (i != y)) {
				z = a[x][i];
				z += a[y][i];
				z *= 500000004;
				z %= MOD;
				a[x][i] = z;
				a[y][i] = z;
			}
		}
		z = a[x][y];
		z += a[y][x];
		z *= 500000004;
		z %= MOD;
		a[x][y] = z;
		a[y][x] = z;
	}
	z = 0;
	f(i, n) {
		f(j, i) {
			z = (z + a[i][j]) % MOD;
		}
	}
	f(i, q)z = (z * 2) % MOD;
	printf("%lld\n", z);
	return 0;
}