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>
#define RI register int
#define CI const int&
using namespace std;
const int N=3005,mod=1e9+7;
int n,q,f[N][N],inv2,mul,a[N],tx[N],ty[N],ans;
inline int quick_pow(int x,int p=mod-2,int mul=1)
{
	for (;p;p>>=1,x=1LL*x*x%mod) if (p&1) mul=1LL*mul*x%mod; return mul;
}
inline int sum(CI x,CI y)
{
	int t=x+y; return t>=mod?t-mod:t;
}
int main()
{
	RI i,j; for (scanf("%d%d",&n,&q),i=1;i<=n;++i) scanf("%d",&a[i]);
	for (i=1;i<=q;++i) scanf("%d%d",&tx[i],&ty[i]); inv2=quick_pow(2);
	for (i=1;i<=n;++i) for (j=1;j<=n;++j) if (i!=j&&a[i]>a[j]) ++f[i][j];
	for (mul=quick_pow(2,q),i=1;i<=q;++i)
	{
		int x=tx[i],y=ty[i]; f[x][y]=f[y][x]=1LL*sum(f[x][y],f[y][x])*inv2%mod;
		for (j=1;j<=n;++j) if (j!=x&&j!=y) f[x][j]=f[y][j]=1LL*sum(f[x][j],f[y][j])*inv2%mod;
		for (j=1;j<=n;++j) if (j!=x&&j!=y) f[j][x]=f[j][y]=1LL*sum(f[j][x],f[j][y])*inv2%mod;
	}
	for (i=1;i<=n;++i) for (j=1;j<=n;++j) if (i<j) ans=sum(ans,1LL*f[i][j]*mul%mod);
	return printf("%d",ans),0;
}