# include <iostream>
# include <cstdio>
# include <cstring>
# include <vector>
# define lson l, m, id<<1
# define rson m+1, r, id<<1|1
using namespace std;
const int maxn = 2e5+30;
const int oo = 0x3f3f3f3f;
int imin[maxn<<2], cost[maxn], dp[maxn];
void build(int l, int r, int id)
{
    if(l == r)
    {
        imin[id] = oo;
        return;
    }
    int m = l+r>>1;
    build(lson);
    build(rson);
    imin[id] = min(imin[id<<1], imin[id<<1|1]);
}
 
int query(int L, int R, int l, int r, int id)//线段树区间维护最小值。
{
    if(L<=l && r<=R) return imin[id];
    int m = l+r>>1, res = oo;
    if(L <= m) res = min(res, query(L, R, lson));
    if(R > m) res = min(res, query(L, R, rson));
    return res;
}
 
void update(int pos, int val, int l, int r, int id)
{
    if(l == r)
    {
        imin[id] = min(imin[id], val);
        return;
    }
    int m = l+r>>1;
    if(pos <= m) update(pos, val, lson);
    else update(pos, val, rson);
    imin[id] = min(imin[id<<1], imin[id<<1|1]);
}
vector<int>v[maxn];
int main()
{
    int n, q, cnt=0;
    scanf("%d",&n);
    build(1, n, 1);
    for(int i=1, j; i<=n; ++i)
    {
        scanf("%d",&j);
        cnt += !j;
        cost[i] = j?1:-1;
    }
    scanf("%d",&q);
    for(int i=0, j, k; i<q; ++i)
    {
        scanf("%d%d",&j,&k);
        v[j].push_back(k);//此处需要枚举左端点。
    }
    memset(dp, oo, sizeof(dp));
    dp[0] = 0;
    for(int i=1; i<=n; ++i)
    {
        for(auto j : v[i])
        {
            int mi = dp[i-1];
            mi = min(mi, query(max(i-1,1), j, 1, n, 1));
            if(mi < dp[j])//选择该区间的情况，★为什么可以直接取mi值呢？因为mi值所在点前面枚举时必然覆盖了i点！可以作为答案更新给dp[j]。
            {
                dp[j] = mi;
                update(j, mi, 1, n, 1);
            }
        }
        dp[i] = min(dp[i], dp[i-1]+cost[i]); //不选择该区间的情况。
    }
    printf("%d\n",dp[n]+cnt);
    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 <iostream>
# include <cstdio>
# include <cstring>
# include <vector>
# define lson l, m, id<<1
# define rson m+1, r, id<<1|1
using namespace std;
const int maxn = 2e5+30;
const int oo = 0x3f3f3f3f;
int imin[maxn<<2], cost[maxn], dp[maxn];
void build(int l, int r, int id)
{
    if(l == r)
    {
        imin[id] = oo;
        return;
    }
    int m = l+r>>1;
    build(lson);
    build(rson);
    imin[id] = min(imin[id<<1], imin[id<<1|1]);
}
 
int query(int L, int R, int l, int r, int id)//线段树区间维护最小值。
{
    if(L<=l && r<=R) return imin[id];
    int m = l+r>>1, res = oo;
    if(L <= m) res = min(res, query(L, R, lson));
    if(R > m) res = min(res, query(L, R, rson));
    return res;
}
 
void update(int pos, int val, int l, int r, int id)
{
    if(l == r)
    {
        imin[id] = min(imin[id], val);
        return;
    }
    int m = l+r>>1;
    if(pos <= m) update(pos, val, lson);
    else update(pos, val, rson);
    imin[id] = min(imin[id<<1], imin[id<<1|1]);
}
vector<int>v[maxn];
int main()
{
    int n, q, cnt=0;
    scanf("%d",&n);
    build(1, n, 1);
    for(int i=1, j; i<=n; ++i)
    {
        scanf("%d",&j);
        cnt += !j;
        cost[i] = j?1:-1;
    }
    scanf("%d",&q);
    for(int i=0, j, k; i<q; ++i)
    {
        scanf("%d%d",&j,&k);
        v[j].push_back(k);//此处需要枚举左端点。
    }
    memset(dp, oo, sizeof(dp));
    dp[0] = 0;
    for(int i=1; i<=n; ++i)
    {
        for(auto j : v[i])
        {
            int mi = dp[i-1];
            mi = min(mi, query(max(i-1,1), j, 1, n, 1));
            if(mi < dp[j])//选择该区间的情况，★为什么可以直接取mi值呢？因为mi值所在点前面枚举时必然覆盖了i点！可以作为答案更新给dp[j]。
            {
                dp[j] = mi;
                update(j, mi, 1, n, 1);
            }
        }
        dp[i] = min(dp[i], dp[i-1]+cost[i]); //不选择该区间的情况。
    }
    printf("%d\n",dp[n]+cnt);
    return 0;
}
```