#include<bits/stdc++.h>
#define X first
#define Y second
using namespace std;
typedef pair<int,int> pii;
const int N = 200005, L = 262144, inf = 1e9;

int n, m, s[N];
pii a[N];

struct minseg {
	int v[2*L];
	void init () {
		for(int i=2*L;i--;) v[i] = inf;
	}
	void upd (int P, int V) {
		P += L;
		while(P) {
			v[P] = min(v[P], V);
			P /= 2;
		}
	}
	int get (int S, int E) {
		S += L; E += L;
		int R = inf;
		while(S <= E) {
			if(S%2 == 1) R = min(R, v[S++]);
			if(E%2 == 0) R = min(R, v[E--]);
			S /= 2; E /= 2;
		}
		return R;
	}
} b, w;

int hmd (int I, int V) {
	if(!V) return s[I];
	return I - s[I];
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) {
		scanf("%d",&s[i]);
		s[i] += s[i-1];
	}
	scanf("%d",&m);
	for(int i=1;i<=m;i++) {
		scanf("%d%d",&a[i].X,&a[i].Y);
	}
	sort(a+1, a+1+m);
	b.init(); w.init();
	w.upd(0, 0);
	for(int i=1;i<=m;i++) {
		int B = b.get(a[i].X-1, a[i].Y-1);
		int W = w.get(0, a[i].X-1);
		b.upd(a[i].Y, B);
		w.upd(a[i].Y, B + hmd(a[i].Y, 1) - hmd(a[i].Y, 0));
		b.upd(a[i].Y, W + hmd(a[i].X-1, 0) - hmd(a[i].X-1, 1));
		w.upd(a[i].Y, W + hmd(a[i].X-1, 0) - hmd(a[i].X-1, 1) + hmd(a[i].Y, 1) - hmd(a[i].Y, 0));
	}
	printf("%d\n", w.get(0,n) + s[n]);
}

### 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<bits/stdc++.h>
#define X first
#define Y second
using namespace std;
typedef pair<int,int> pii;
const int N = 200005, L = 262144, inf = 1e9;

int n, m, s[N];
pii a[N];

struct minseg {
	int v[2*L];
	void init () {
		for(int i=2*L;i--;) v[i] = inf;
	}
	void upd (int P, int V) {
		P += L;
		while(P) {
			v[P] = min(v[P], V);
			P /= 2;
		}
	}
	int get (int S, int E) {
		S += L; E += L;
		int R = inf;
		while(S <= E) {
			if(S%2 == 1) R = min(R, v[S++]);
			if(E%2 == 0) R = min(R, v[E--]);
			S /= 2; E /= 2;
		}
		return R;
	}
} b, w;

int hmd (int I, int V) {
	if(!V) return s[I];
	return I - s[I];
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) {
		scanf("%d",&s[i]);
		s[i] += s[i-1];
	}
	scanf("%d",&m);
	for(int i=1;i<=m;i++) {
		scanf("%d%d",&a[i].X,&a[i].Y);
	}
	sort(a+1, a+1+m);
	b.init(); w.init();
	w.upd(0, 0);
	for(int i=1;i<=m;i++) {
		int B = b.get(a[i].X-1, a[i].Y-1);
		int W = w.get(0, a[i].X-1);
		b.upd(a[i].Y, B);
		w.upd(a[i].Y, B + hmd(a[i].Y, 1) - hmd(a[i].Y, 0));
		b.upd(a[i].Y, W + hmd(a[i].X-1, 0) - hmd(a[i].X-1, 1));
		w.upd(a[i].Y, W + hmd(a[i].X-1, 0) - hmd(a[i].X-1, 1) + hmd(a[i].Y, 1) - hmd(a[i].Y, 0));
	}
	printf("%d\n", w.get(0,n) + s[n]);
}

```