#include<cstdio>
#include<cstring>
#include<iostream>
#include<stdlib.h>
#include<ctime>
#include<string>
#include<cmath>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define LL long long
#define FOR(i,a,b) for (int i=a;i<=b;i++)
#define FORD(i,a,b) for (int i=a;i>=b;i--)
using namespace std;
typedef pair<int,int> pa;
void getint(int &v){
    char ch,fu=0;
    for(ch='*'; (ch<'0'||ch>'9')&&ch!='-'; ch=getchar());
    if(ch=='-') fu=1, ch=getchar();
    for(v=0; ch>='0'&&ch<='9'; ch=getchar()) v=v*10+ch-'0';
    if(fu) v=-v;
}
const int INF=1e9+10;
int n,a[500010],ans,Q,f[500010],s[500010],cnt,T[2][1000010];
pa q[500010];
void upd(int tp,int l,int r,int rt,int x,int y){
	if (l==r){
		T[tp][rt]=max(T[tp][rt],y);
		return;
	}
	int m=l+r>>1;
	if (x<=m) upd(tp,l,m,rt<<1,x,y);
	else upd(tp,m+1,r,rt<<1|1,x,y);
	T[tp][rt]=max(T[tp][rt<<1],T[tp][rt<<1|1]);
}
int que(int tp,int l,int r,int rt,int L,int R){
	if (l>=L && r<=R) return T[tp][rt];
	int m=l+r>>1,t=-INF;
	if (L<=m) t=max(t,que(tp,l,m,rt<<1,L,R));
	if (R>m) t=max(t,que(tp,m+1,r,rt<<1|1,L,R));
	return t;
}
int main(){
	scanf("%d",&n);
	FOR(i,1,n){
		scanf("%d",&a[i]);
		if (a[i]==0) ++cnt,a[i]=-1;
		s[i]=s[i-1]+a[i];
	}
	scanf("%d",&Q);
	FOR(i,1,Q) scanf("%d%d",&q[i].fi,&q[i].se);
	sort(q+1,q+Q+1);
	FOR(i,0,1) FOR(j,0,1000000) T[i][j]=-INF;
	FOR(j,1,Q){
		f[j]=s[q[j].se]-s[q[j].fi-1];
		if (1<=q[j].fi-1) f[j]=max(f[j],que(0,1,n,1,1,q[j].fi-1)+s[q[j].se]-s[q[j].fi-1]);
		if (q[j].fi<=q[j].se-1) f[j]=max(f[j],que(1,1,n,1,q[j].fi,q[j].se-1)+s[q[j].se]);
		upd(0,1,n,1,q[j].se,f[j]);
		upd(1,1,n,1,q[j].se,f[j]-s[q[j].se]);
		ans=max(ans,f[j]);
	}
	cout<<n-ans-cnt<<endl;
	return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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

### Response

```cpp
#include<cstdio>
#include<cstring>
#include<iostream>
#include<stdlib.h>
#include<ctime>
#include<string>
#include<cmath>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define LL long long
#define FOR(i,a,b) for (int i=a;i<=b;i++)
#define FORD(i,a,b) for (int i=a;i>=b;i--)
using namespace std;
typedef pair<int,int> pa;
void getint(int &v){
    char ch,fu=0;
    for(ch='*'; (ch<'0'||ch>'9')&&ch!='-'; ch=getchar());
    if(ch=='-') fu=1, ch=getchar();
    for(v=0; ch>='0'&&ch<='9'; ch=getchar()) v=v*10+ch-'0';
    if(fu) v=-v;
}
const int INF=1e9+10;
int n,a[500010],ans,Q,f[500010],s[500010],cnt,T[2][1000010];
pa q[500010];
void upd(int tp,int l,int r,int rt,int x,int y){
	if (l==r){
		T[tp][rt]=max(T[tp][rt],y);
		return;
	}
	int m=l+r>>1;
	if (x<=m) upd(tp,l,m,rt<<1,x,y);
	else upd(tp,m+1,r,rt<<1|1,x,y);
	T[tp][rt]=max(T[tp][rt<<1],T[tp][rt<<1|1]);
}
int que(int tp,int l,int r,int rt,int L,int R){
	if (l>=L && r<=R) return T[tp][rt];
	int m=l+r>>1,t=-INF;
	if (L<=m) t=max(t,que(tp,l,m,rt<<1,L,R));
	if (R>m) t=max(t,que(tp,m+1,r,rt<<1|1,L,R));
	return t;
}
int main(){
	scanf("%d",&n);
	FOR(i,1,n){
		scanf("%d",&a[i]);
		if (a[i]==0) ++cnt,a[i]=-1;
		s[i]=s[i-1]+a[i];
	}
	scanf("%d",&Q);
	FOR(i,1,Q) scanf("%d%d",&q[i].fi,&q[i].se);
	sort(q+1,q+Q+1);
	FOR(i,0,1) FOR(j,0,1000000) T[i][j]=-INF;
	FOR(j,1,Q){
		f[j]=s[q[j].se]-s[q[j].fi-1];
		if (1<=q[j].fi-1) f[j]=max(f[j],que(0,1,n,1,1,q[j].fi-1)+s[q[j].se]-s[q[j].fi-1]);
		if (q[j].fi<=q[j].se-1) f[j]=max(f[j],que(1,1,n,1,q[j].fi,q[j].se-1)+s[q[j].se]);
		upd(0,1,n,1,q[j].se,f[j]);
		upd(1,1,n,1,q[j].se,f[j]-s[q[j].se]);
		ans=max(ans,f[j]);
	}
	cout<<n-ans-cnt<<endl;
	return 0;
}

```