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<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x); i<=(y); i++)
#define ll long long
#define ld long double
#define inf 1000000000
using namespace std;
#define N 200005
#define lc (o<<1)
#define rc (o<<1|1)
int n,m,b[N],sum,mi[N<<2],tg[N<<2];
struct node{ int l,r; }a[N];
const bool cmp(const node &x,const node &y){ return x.l<y.l; }
void down(int o){
	if (!tg[o]) return;
	mi[lc]+=tg[o]; mi[rc]+=tg[o]; tg[o]=0;
}
void up(int o){ mi[o]=min(mi[lc],mi[rc]); }
int qrymin(int o,int l,int r,int x,int y){
	if (x<=l && r<=y) return mi[o];
	down(o); int mid=l+r>>1,ret=inf;
	if (x<=mid) ret=min(ret,qrymin(lc,l,mid,x,y));
	if (y>mid) ret=min(ret,qrymin(rc,mid+1,r,x,y));
	return ret;
}
void modify(int o,int l,int r,int x,int val){
	if (l==r){ mi[o]=val; return; }
	down(o); int mid=l+r>>1;
	if (x<=mid) modify(lc,l,mid,x,val);
	else modify(rc,mid+1,r,x,val); up(o);
}
void add(int o,int l,int r,int x,int y,int val){
	if (x<=l && r<=y){ tg[o]+=val,mi[o]+=val; return; }
	down(o); int mid=l+r>>1;
	if (x<=mid) add(lc,l,mid,x,y,val);
	if (y>mid) add(rc,mid+1,r,x,y,val); up(o);
}
int main(){
	scanf("%d",&n);
	rep (i,1,n) scanf("%d",&b[i]),sum+=!b[i];
	scanf("%d",&m);
	rep (i,1,m) scanf("%d%d",&a[i].l,&a[i].r);
	sort(a+1,a+1+m,cmp); int po=1;
	memset(mi,127,sizeof(mi)); modify(1,0,n,0,0);
	rep (i,1,n){
		//f[i][j]表示前i个数已经确定，最后一个1在第j位，最小的(0,1)-(0,0)。一位一位确定。
		while (po<=m && a[po].l<=i){
			modify(1,0,n,a[po].r,qrymin(1,0,n,0,a[po].r));//f[i][r]=min{f[i-1][k]}(k<=r);
			po++;
		}
		add(1,0,n,0,i-1,b[i]?1:-1);
	}
	printf("%d\n",mi[1]+sum);
	return 0;
}