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 <bits/stdc++.h>

typedef long long int int64;
static const int64 mod = 1000000007LL;
int64 dp[3003][3003];
int64 temp[2][3003];
int a[3003];
int n,q;
int main()
{
	scanf("%d%d",&n,&q);
	for(int i = 0;i < n;++i) {
		scanf("%d",&a[i]);
	}
	for(int i = 0;i < n;++i) {
		for(int j = 0;j < n;++j) {
			dp[i][j] = a[i] > a[j] ? 1 : 0;
		}
	}
	int64 inv2 = mod - mod / 2;
	int64 val = 1;
	for(int i = 0;i < q;++i) {
		int x,y;
		scanf("%d%d",&x,&y);--x,--y;
		for(int j = 0;j < n;++j) {
			// dp[x][j] = dp[y][j] = (dp[x][j]+dp[y][j])/2
			// swap(dp[x][y], dp[y][x])
			// dp[x][y] = dp[y][x] = (dp[x][y]+dp[y][x])/2
			temp[0][j] = (dp[x][j] + dp[y][j]) * inv2 % mod;
			temp[1][j] = (dp[j][x] + dp[j][y]) * inv2 % mod;
		}
		for(int j = 0;j < n;++j) {
			if(j != x && j != y) {
				dp[x][j] = temp[0][j];
				dp[j][x] = temp[1][j];
				dp[y][j] = temp[0][j];
				dp[j][y] = temp[1][j];
			}
		}
		int64 t2 = (dp[x][y] + dp[y][x]) * inv2 % mod;
		dp[x][y] = dp[y][x] = t2;
		val = val * 2 % mod;
	}
	int64 ans = 0;
	for(int i = 0;i < n;++i) {
		for(int j = i+1;j < n;++j) {
			ans = (ans + dp[i][j]) % mod;
		}
	}
	ans = ans * val % mod;
	printf("%lld\n", ans);
}