#include <bits/stdc++.h>
using namespace std;
using lint = long long;
using pi = pair<int, int>;
const int MAXN = 200005;
const int MAXT = 530000;

struct seg{
	int lim, tree[MAXT];
	void init(int n){
		memset(tree, 0x3f, sizeof(tree));
		for(lim = 1; lim <= n; lim <<= 1);
	}
	void update(int x, int v){
		x += lim;
		tree[x] = min(tree[x], v);
		while(x > 1){
			x >>= 1;
			tree[x] = min(tree[2*x], tree[2*x+1]);
		}
	}
	int query(int s, int e){
		 s+=lim;
		 e+=lim;
		 int ret=1e9;
		 while(s < e){
		 	 if(s%2 == 1) ret = min(ret, tree[s++]);
		 	 if(e%2 == 0) ret = min(ret, tree[e--]);
		 	 s >>= 1;
		 	 e >>= 1;
		 }
		 if(s == e) ret= min(ret, tree[s]);
		 return ret;
	}
}seg;

int n, q;
vector<int> in[MAXN];
vector<int> out[MAXN];
pi intv[MAXN];
int d1[MAXN], d2[MAXN];
int s[MAXN], b[MAXN];

int main(){
	scanf("%d",&n);
	for(int i=1; i<=n; i++){
		scanf("%d",&b[i]);
		s[i] = s[i-1] + (1 - b[i]);
	}
	scanf("%d",&q);
	for(int i=0; i<q; i++){
		scanf("%d %d",&intv[i].first,&intv[i].second);
	}
	sort(intv, intv + q, [&](const pi &a, const pi &b){
		return a.second < b.second;
	});
	for(int i=0; i<q; i++){
		in[intv[i].first].push_back(i);
		out[intv[i].second + 1].push_back(i);
	}
	seg.init(n + 5);
	for(int i=n+1; i; i--){
		if(i < n + 1){
			d1[i] = d1[i + 1] + b[i];
			for(auto &j : in[i]) d1[i] = min(d1[i], d2[j] + s[intv[j].second] - s[intv[j].first - 1]);
		}
		for(auto &j : out[i]){
			d2[j] = min(seg.query(intv[j].first + 1, intv[j].second) - s[intv[j].second], d1[i]);
		}
		for(auto &j : out[i]){
			seg.update(intv[j].first, d2[j] + s[intv[j].second]);
		}
	}
	cout << d1[1] << endl;
}


### 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>
using namespace std;
using lint = long long;
using pi = pair<int, int>;
const int MAXN = 200005;
const int MAXT = 530000;

struct seg{
	int lim, tree[MAXT];
	void init(int n){
		memset(tree, 0x3f, sizeof(tree));
		for(lim = 1; lim <= n; lim <<= 1);
	}
	void update(int x, int v){
		x += lim;
		tree[x] = min(tree[x], v);
		while(x > 1){
			x >>= 1;
			tree[x] = min(tree[2*x], tree[2*x+1]);
		}
	}
	int query(int s, int e){
		 s+=lim;
		 e+=lim;
		 int ret=1e9;
		 while(s < e){
		 	 if(s%2 == 1) ret = min(ret, tree[s++]);
		 	 if(e%2 == 0) ret = min(ret, tree[e--]);
		 	 s >>= 1;
		 	 e >>= 1;
		 }
		 if(s == e) ret= min(ret, tree[s]);
		 return ret;
	}
}seg;

int n, q;
vector<int> in[MAXN];
vector<int> out[MAXN];
pi intv[MAXN];
int d1[MAXN], d2[MAXN];
int s[MAXN], b[MAXN];

int main(){
	scanf("%d",&n);
	for(int i=1; i<=n; i++){
		scanf("%d",&b[i]);
		s[i] = s[i-1] + (1 - b[i]);
	}
	scanf("%d",&q);
	for(int i=0; i<q; i++){
		scanf("%d %d",&intv[i].first,&intv[i].second);
	}
	sort(intv, intv + q, [&](const pi &a, const pi &b){
		return a.second < b.second;
	});
	for(int i=0; i<q; i++){
		in[intv[i].first].push_back(i);
		out[intv[i].second + 1].push_back(i);
	}
	seg.init(n + 5);
	for(int i=n+1; i; i--){
		if(i < n + 1){
			d1[i] = d1[i + 1] + b[i];
			for(auto &j : in[i]) d1[i] = min(d1[i], d2[j] + s[intv[j].second] - s[intv[j].first - 1]);
		}
		for(auto &j : out[i]){
			d2[j] = min(seg.query(intv[j].first + 1, intv[j].second) - s[intv[j].second], d1[i]);
		}
		for(auto &j : out[i]){
			seg.update(intv[j].first, d2[j] + s[intv[j].second]);
		}
	}
	cout << d1[1] << endl;
}


```