Problem: You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

Solution (in CPP):

Solution:
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#define MN 200005
using namespace std;
int n,q,cnt,T[MN<<2],f[MN],c[MN];
vector<int> v[MN];
void update(int l,int r,int a,int k,int ad){
	if(l==r)return (void)(T[k]=min(T[k],ad));int mid=(l+r)>>1;
	if(a<=mid)update(l,mid,a,k<<1,ad);else update(mid+1,r,a,k<<1|1,ad);T[k]=min(T[k<<1],T[k<<1|1]);
}
int que(int l,int r,int a,int b,int k){
	if(l==a&&r==b)return T[k];int mid=(l+r)>>1;
	if(b<=mid)return que(l,mid,a,b,k<<1);if(a>mid)return que(mid+1,r,a,b,k<<1|1);
	return min(que(l,mid,a,mid,k<<1),que(mid+1,r,mid+1,b,k<<1|1));
}
int main(){
	memset(T,0x3f,sizeof(T));memset(f,0x3f,sizeof(f));
	scanf("%d",&n);int x,y;
	for(int i=1;i<=n;i++)scanf("%d",&x),c[i]=x?1:-1,cnt+=!x;
	scanf("%d",&q);
	for(int i=1;i<=q;i++)scanf("%d%d",&x,&y),v[x].push_back(y);f[0]=0;
	for(int i=1;i<=n;i++){
		for(int j=0;j<v[i].size();j++){
			int tmp=min(f[i-1],que(1,n,max(i-1,1),v[i][j],1));
			if(tmp<f[v[i][j]])f[v[i][j]]=tmp,update(1,n,v[i][j],1,tmp);
		}f[i]=min(f[i],f[i-1]+c[i]);
	}printf("%d\n",f[n]+cnt);
}