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> 
#include <cstdlib> 
#include <cstring> 
#include <cmath> 
#include <algorithm>  
using namespace std; 
const int MAX_N = 3e3 + 5; 
const int Mod = 1e9 + 7; 
int fpow(int x, int y) {
	int res = 1; 
	while (y) { 
		if (y & 1) res = 1ll * res * x % Mod; 
		x = 1ll * x * x % Mod; 
		y >>= 1; 
	} 
	return res; 
} 
const int inv = fpow(2, Mod - 2); 
int N, Q, a[MAX_N]; 
int f[MAX_N][MAX_N]; 
int main () { 
	scanf("%d %d", &N, &Q); 
	for (int i = 1; i <= N; i++) scanf("%d", a + i); 
	for (int i = 1; i <= N; i++) 
		for (int j = 1; j <= N; j++) f[i][j] = a[i] < a[j]; 
	for (int i = 1; i <= Q; i++) { 
		int x, y; scanf("%d %d", &x, &y); 
		f[x][y] = f[y][x] = 1ll * (f[x][y] + f[y][x]) * inv % Mod; 
		for (int j = 1; j <= N; j++) { 
			if (j == x || j == y) continue; 
			f[x][j] = f[y][j] = 1ll * (f[x][j] + f[y][j]) * inv % Mod; 
			f[j][x] = f[j][y] = 1ll * (f[j][x] + f[j][y]) * inv % Mod; 
		} 
	} 
	int ans = 0; 
	for (int i = 1; i <= N; i++) 
		for (int j = 1; j < i; j++) 
			ans = (ans + f[i][j]) % Mod; 
	printf("%lld\n", 1ll * ans * fpow(2, Q) % Mod); 
    return 0;  
}