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 <iostream>
#include <cstdio>
#define ll long long
using namespace std;
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
const int mod = 1e9 + 7,N = 4e3;
int n,m; ll Ans = 1,a[N],f[N][N];
int main()
{
	n = read(); m = read();
	for(int i = 1;i <= n;i ++) a[i] = read();
	for(int i = 1;i <= n;i ++) for(int j = 1;j <= n;j ++) f[i][j] = a[i] > a[j];
	int x,y,inv2 = (mod + 1) / 2;
	while(m --)
	{
		Ans = (Ans << 1) % mod;
		x = read(); y = read();
		f[x][y] = f[y][x] = 1ll * ((f[x][y] + f[y][x]) % mod + mod) % mod * inv2 % mod;
		for(int i = 1;i <= n;i ++)
		{
			if(i == x || i == y) continue;
			f[x][i] = f[y][i] = 1ll * ((f[x][i] + f[y][i]) % mod + mod) % mod * inv2 % mod;
			f[i][x] = f[i][y] = 1ll * ((f[i][x] + f[i][y]) % mod + mod) % mod * inv2 % mod;
		}
	}
	int tot = 0;
	for(int i = 1;i <= n;i ++) for(int j = i + 1;j <= n;j ++) tot = (tot + f[i][j]) % mod;
	cout << ((1ll * Ans * tot % mod) + mod) % mod;
}