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<cstring>
#include<algorithm>
#include<vector>
#define inf 0x3f3f3f3f
#define MM (1<<20)
#define MN 200005
using namespace std;
inline int in(){
    int x=0;bool f=0; char c;
    for (;(c=getchar())<'0'||c>'9';f=c=='-');
    for (x=c-'0';(c=getchar())>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0');
    return f?-x:x;
}
vector<int> v[MN];
int mn[MM],f[MN],b[MN],sum[MN];
int n,l,r,M,q;
inline void update(int x,int v){
	for (x+=M;x;x>>=1) mn[x]=min(mn[x],v);
}
inline int query (int l,int r){
	int res=inf;
	for (l+=M-1,r+=M+1;l^r^1;l>>=1,r>>=1){
		if (~l&1) res=min(res,mn[l^1]);
		if ( r&1) res=min(res,mn[r^1]);
	}return res;
}
int main()
{
	n=in();memset(f,0x3f,sizeof(f));f[0]=0;
	for (int i=1;i<=n;++i) b[i]=in(),sum[i]=sum[i-1]+1-b[i];
	memset(mn,0x3f,sizeof(mn));
	for (M=1;M<n+2;M<<=1);q=in();
	for (int i=1;i<=q;++i){
		l=in();r=in();v[l-1].push_back(r);
	}
	for (int i=0;i<n;++i){
		f[i+1]=min(f[i+1],f[i]+b[i+1]);
		int siz=v[i].size();
		for (int j=0;j<siz;++j){
			int dis=min(f[i]+sum[v[i][j]]-sum[i],sum[v[i][j]]+query(i+1,v[i][j]));
			f[v[i][j]]=min(f[v[i][j]],dis);
			update(v[i][j]+1,dis-sum[v[i][j]]);		
		}
	}printf("%d",f[n]);return 0;
}