#include<bits/stdc++.h>
#define rep(i, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
using std::cin; using std::cout;
const int maxn = 200200;
typedef long long ll;
const int mod = 998244353;
inline ll pow(ll a,int b,int ans = 1) {
	for(;b;b >>= 1,a = a * a % mod) if(b & 1)
		ans = ans * a % mod;
	return ans;
}
inline ll inverse(int x){ return pow(x, mod - 2); }
int n, q;
int max[maxn << 2];
inline void upt(int pos,int v,int cur = 1,int l = 0,int r = n) {
	if(l == r) return void(max[cur] = v);
	int mid = l + r >> 1;
	pos <= mid ? upt(pos, v, cur << 1, l, mid) : upt(pos, v, cur << 1 | 1, mid + 1, r);
	max[cur] = std::max(max[cur << 1], max[cur << 1 | 1]);
}
inline int qry(int ql,int qr,int cur = 1,int l = 0,int r = n) {
	if(ql <= l && r <= qr) return max[cur];
	int mid = l + r >> 1, ans = -1e9;
	if(ql <= mid) ans = std::max(ans, qry(ql, qr, cur << 1, l, mid));
	if(mid < qr) ans = std::max(ans, qry(ql, qr, cur << 1 | 1, mid + 1, r));
	return ans;
}
std::vector<int> v[maxn];
int b[maxn];
int dp[maxn];
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n;
	rep(i, 1, n) {
		cin >> b[i];
		b[i] += b[i - 1];
	}
	cin >> q;
	rep(i, 1, q) {
		int l, r;
		cin >> l >> r;
		v[l].push_back(r);
	}
	rep(i, 1, n) dp[i] = -1e9, upt(i, -1e9);
	rep(i, 1, n) {
		for(int x : v[i]) {
			int v = std::max(dp[i - 1], qry(i - 1, x));
			if(v > dp[x]) {
				dp[x] = v;
				upt(x, v);
			}
		}
		dp[i] = std::max(dp[i], dp[i - 1] + (b[i] == b[i - 1] ? 1 : -1));
	}
	cout << n - b[n] - dp[n] << '\n';
}
### Prompt

Please provide a cpp coded solution 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 rep(i, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
using std::cin; using std::cout;
const int maxn = 200200;
typedef long long ll;
const int mod = 998244353;
inline ll pow(ll a,int b,int ans = 1) {
	for(;b;b >>= 1,a = a * a % mod) if(b & 1)
		ans = ans * a % mod;
	return ans;
}
inline ll inverse(int x){ return pow(x, mod - 2); }
int n, q;
int max[maxn << 2];
inline void upt(int pos,int v,int cur = 1,int l = 0,int r = n) {
	if(l == r) return void(max[cur] = v);
	int mid = l + r >> 1;
	pos <= mid ? upt(pos, v, cur << 1, l, mid) : upt(pos, v, cur << 1 | 1, mid + 1, r);
	max[cur] = std::max(max[cur << 1], max[cur << 1 | 1]);
}
inline int qry(int ql,int qr,int cur = 1,int l = 0,int r = n) {
	if(ql <= l && r <= qr) return max[cur];
	int mid = l + r >> 1, ans = -1e9;
	if(ql <= mid) ans = std::max(ans, qry(ql, qr, cur << 1, l, mid));
	if(mid < qr) ans = std::max(ans, qry(ql, qr, cur << 1 | 1, mid + 1, r));
	return ans;
}
std::vector<int> v[maxn];
int b[maxn];
int dp[maxn];
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n;
	rep(i, 1, n) {
		cin >> b[i];
		b[i] += b[i - 1];
	}
	cin >> q;
	rep(i, 1, q) {
		int l, r;
		cin >> l >> r;
		v[l].push_back(r);
	}
	rep(i, 1, n) dp[i] = -1e9, upt(i, -1e9);
	rep(i, 1, n) {
		for(int x : v[i]) {
			int v = std::max(dp[i - 1], qry(i - 1, x));
			if(v > dp[x]) {
				dp[x] = v;
				upt(x, v);
			}
		}
		dp[i] = std::max(dp[i], dp[i - 1] + (b[i] == b[i - 1] ? 1 : -1));
	}
	cout << n - b[n] - dp[n] << '\n';
}
```