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<cctype>
#include<cstring>
#include<algorithm>
const int maxn=500010;
int n,m;
bool v[maxn];
struct Interval {
	int l,r;
	bool operator < (const Interval &t) const{
		return l==t.l?r>t.r:l<t.l;
	}
}t[maxn];
int s[maxn<<2],o;
int ad;
int In() {
	register char c;
	for(;c=getchar(),c!='-'&&!isdigit(c););
	register bool f=c=='-';
	register int s=f?0:c-'0';
	for(;c=getchar(),isdigit(c);)s=s*10+c-'0';
	return f?-s:s;
}
int dmin(register int x,register int y) {
	return x<y?x:y;
}
void mdf(register int x,register int v) {
	if(s[x+=o]<=v) return;
	for(s[x]=v;x>>=1;s[x]=dmin(s[x<<1],s[x<<1|1]));
}
int qry(register int l,register int r) {
	register int z=0x3f3f3f3f;
	for(l+=o-1,r+=o+1;l^r^1;l>>=1,r>>=1)~l&1?z=dmin(z,s[l^1]):0,r&1?z=dmin(z,s[r^1]):0;
	return z;
}
int main() {
	n=In();
	for(register int i=1;i<=n;++i)ad+=!(v[i]=In());
	m=In();
	for(register int i=1;i<=m;++i)t[i]=(Interval){In(),In()};
	std::sort(t+1,t+m+1);
	memset(s,63,sizeof s);
	for(o=1;o<=n+1;o<<=1);
	mdf(0,0);
	for(register int i=1,j=1;i<=n;++i) {
		for(;j<=m&&t[j].l==i;++j)mdf(t[j].r,qry(i-1,t[j].r)); 
		mdf(i,s[i-1+o]+(v[i]?1:-1));
	}
	printf("%d\n",s[n+o]+ad);
	return 0;
}