#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10;
vector<int> g[maxn];
int vis[maxn];
stack<int> S;
int p[maxn];
long long sum[maxn];
void dfs(int u, int fa) {
  S.push(u);
  vis[u] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v != fa) {
      if (vis[v] == 0)
        dfs(v, u);
      else if (vis[v] == 1) {
        int Max = u, Min = u;
        while (!S.empty()) {
          int w = S.top();
          S.pop();
          Max = max(Max, w);
          Min = min(Min, w);
          if (w == v) break;
        }
        p[Min] = Max;
      }
    }
  }
  if (!S.empty() && S.top() == u) S.pop();
  vis[u] = 2;
}
int main() {
  int n, m, q;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  memset(vis, 0, sizeof vis);
  for (int i = 1; i <= n + 1; i++) p[i] = n + 1;
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) dfs(i, -1);
  }
  sum[n + 1] = 0;
  for (int i = n; i >= 1; i--) {
    p[i] = min(p[i], p[i + 1]);
    sum[i] = p[i] - i + sum[i + 1];
  }
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int L = l, R = r + 1;
    while (L < R) {
      int m = L + R >> 1;
      if (p[m] <= r)
        L = m + 1;
      else
        R = m;
    }
    printf("%I64d\n", sum[l] - sum[L] + 1LL * (r - L + 2) * (r - L + 1) / 2);
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n. 

You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.

The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.

The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.

The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.

Output

Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.

Examples

Input

6 6
1 2
2 3
3 1
4 5
5 6
6 4
3
1 3
4 6
1 6


Output

5
5
14


Input

8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8


Output

27
8
19

Note

The first example is shown on the picture below:

<image>

For the first query, all subsegments of [1; 3], except this segment itself, are suitable.

For the first query, all subsegments of [4; 6], except this segment itself, are suitable.

For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].

The second example is shown on the picture below:

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10;
vector<int> g[maxn];
int vis[maxn];
stack<int> S;
int p[maxn];
long long sum[maxn];
void dfs(int u, int fa) {
  S.push(u);
  vis[u] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v != fa) {
      if (vis[v] == 0)
        dfs(v, u);
      else if (vis[v] == 1) {
        int Max = u, Min = u;
        while (!S.empty()) {
          int w = S.top();
          S.pop();
          Max = max(Max, w);
          Min = min(Min, w);
          if (w == v) break;
        }
        p[Min] = Max;
      }
    }
  }
  if (!S.empty() && S.top() == u) S.pop();
  vis[u] = 2;
}
int main() {
  int n, m, q;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  memset(vis, 0, sizeof vis);
  for (int i = 1; i <= n + 1; i++) p[i] = n + 1;
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) dfs(i, -1);
  }
  sum[n + 1] = 0;
  for (int i = n; i >= 1; i--) {
    p[i] = min(p[i], p[i + 1]);
    sum[i] = p[i] - i + sum[i + 1];
  }
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int L = l, R = r + 1;
    while (L < R) {
      int m = L + R >> 1;
      if (p[m] <= r)
        L = m + 1;
      else
        R = m;
    }
    printf("%I64d\n", sum[l] - sum[L] + 1LL * (r - L + 2) * (r - L + 1) / 2);
  }
  return 0;
}

```