#include <bits/stdc++.h>
using namespace std;
inline int read() {
  register int res = 0, c, d = 0;
  while (c = getchar(), c < '0' || c > '9')
    ;
  do {
    res = (res << 3) + (res << 1) + (c ^ 48);
  } while (c = getchar(), c >= '0' && c <= '9');
  return res;
}
const int maxn = 300010;
vector<int> edge[maxn];
stack<int> S;
bool vis[maxn];
int low[maxn], dfn[maxn], dep;
int To[maxn], cnt, xl[maxn], xr[maxn];
long long Sum[maxn];
void dfs(int x, int from) {
  low[x] = dfn[x] = dep++;
  vis[x] = 1;
  S.push(x);
  for (auto i : edge[x])
    if (i != from) {
      if (!vis[i]) {
        dfs(i, x);
        low[x] = min(low[x], low[i]);
      } else
        low[x] = min(low[x], dfn[i]);
    }
  if (low[x] == dfn[x]) {
    xl[cnt] = xr[cnt] = x;
    do {
      xl[cnt] = min(xl[cnt], S.top());
      xr[cnt] = max(xr[cnt], S.top());
      if (S.top() == x) break;
      S.pop();
    } while (1);
    S.pop();
    cnt++;
  }
}
int main() {
  int N = read(), M = read();
  for (int i = 1; i <= M; i++) {
    int l = read(), r = read();
    edge[l].push_back(r);
    edge[r].push_back(l);
  }
  for (int i = 1; i <= N; i++) To[i] = N;
  for (int i = 1; i <= N; i++)
    if (!vis[i]) {
      dfs(i, 0);
    }
  for (int i = 0; i < cnt; i++)
    if (xl[i] != xr[i]) {
      To[xl[i]] = xr[i] - 1;
    }
  for (int i = N - 1; i; i--) {
    To[i] = min(To[i], To[i + 1]);
  }
  for (int i = 1; i <= N; i++) {
    Sum[i] = Sum[i - 1] + (To[i] - i + 1);
  }
  int Q = read();
  for (int i = 1; i <= Q; i++) {
    int l = read(), r = read(), L = l - 1, R = r + 1;
    while (L + 1 < R) {
      int mid = L + R >> 1;
      if (To[mid] > r)
        R = mid;
      else
        L = mid;
    }
    long long ans = Sum[L] - Sum[l - 1] + 1ll * (r - L) * (r - L + 1) / 2;
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
inline int read() {
  register int res = 0, c, d = 0;
  while (c = getchar(), c < '0' || c > '9')
    ;
  do {
    res = (res << 3) + (res << 1) + (c ^ 48);
  } while (c = getchar(), c >= '0' && c <= '9');
  return res;
}
const int maxn = 300010;
vector<int> edge[maxn];
stack<int> S;
bool vis[maxn];
int low[maxn], dfn[maxn], dep;
int To[maxn], cnt, xl[maxn], xr[maxn];
long long Sum[maxn];
void dfs(int x, int from) {
  low[x] = dfn[x] = dep++;
  vis[x] = 1;
  S.push(x);
  for (auto i : edge[x])
    if (i != from) {
      if (!vis[i]) {
        dfs(i, x);
        low[x] = min(low[x], low[i]);
      } else
        low[x] = min(low[x], dfn[i]);
    }
  if (low[x] == dfn[x]) {
    xl[cnt] = xr[cnt] = x;
    do {
      xl[cnt] = min(xl[cnt], S.top());
      xr[cnt] = max(xr[cnt], S.top());
      if (S.top() == x) break;
      S.pop();
    } while (1);
    S.pop();
    cnt++;
  }
}
int main() {
  int N = read(), M = read();
  for (int i = 1; i <= M; i++) {
    int l = read(), r = read();
    edge[l].push_back(r);
    edge[r].push_back(l);
  }
  for (int i = 1; i <= N; i++) To[i] = N;
  for (int i = 1; i <= N; i++)
    if (!vis[i]) {
      dfs(i, 0);
    }
  for (int i = 0; i < cnt; i++)
    if (xl[i] != xr[i]) {
      To[xl[i]] = xr[i] - 1;
    }
  for (int i = N - 1; i; i--) {
    To[i] = min(To[i], To[i + 1]);
  }
  for (int i = 1; i <= N; i++) {
    Sum[i] = Sum[i - 1] + (To[i] - i + 1);
  }
  int Q = read();
  for (int i = 1; i <= Q; i++) {
    int l = read(), r = read(), L = l - 1, R = r + 1;
    while (L + 1 < R) {
      int mid = L + R >> 1;
      if (To[mid] > r)
        R = mid;
      else
        L = mid;
    }
    long long ans = Sum[L] - Sum[l - 1] + 1ll * (r - L) * (r - L + 1) / 2;
    printf("%lld\n", ans);
  }
  return 0;
}

```