#include <bits/stdc++.h>
using namespace std;
int N, M, id[300011], bcc = 1;
int num[300011], low[300011], st[300011];
int Q, q[300011][2];
long long csum, inc1, inc2, change[300011], ans[300011];
priority_queue<pair<int, int> > pq1;
priority_queue<pair<int, int> > pq2;
vector<int> adj[300011];
stack<int> stdfs;
priority_queue<pair<int, int> > rnge;
struct segtree {
  long long data[4 * 300011], lazy[4 * 300011];
  void push(int w, int L, int R) {
    if (!lazy[w])
      return;
    else if (L != R) {
      lazy[w * 2] += lazy[w];
      lazy[w * 2 + 1] += lazy[w];
    }
    data[w] += lazy[w] * 1LL * (R - L + 1);
    lazy[w] = 0;
  }
  void update(int w, int L, int R, int a, int b, int v) {
    push(w, L, R);
    if (b < L || a > R)
      return;
    else if (a <= L && b >= R) {
      lazy[w] += v;
      push(w, L, R);
    } else {
      update(w * 2, L, (L + R) / 2, a, b, v);
      update(w * 2 + 1, (L + R) / 2 + 1, R, a, b, v);
      data[w] = data[w * 2] + data[w * 2 + 1];
    }
  }
  long long query(int w, int L, int R, int a, int b) {
    push(w, L, R);
    if (b < L || a > R)
      return 0;
    else if (a <= L && b >= R)
      return data[w];
    else {
      long long q1 = query(w * 2, L, (L + R) / 2, a, b);
      long long q2 = query(w * 2 + 1, (L + R) / 2 + 1, R, a, b);
      return q1 + q2;
    }
  }
} tree;
void dfs(int x, int p) {
  static int cnt = 0;
  bool isAp = false;
  stdfs.push(x);
  num[x] = low[x] = (++cnt);
  for (int i = 0; i < (int)adj[x].size(); i++) {
    int nxt = adj[x][i];
    if (nxt == p) continue;
    if (!num[nxt]) {
      dfs(nxt, x);
      isAp |= (low[nxt] >= num[x]);
    }
    low[x] = min(low[x], low[nxt]);
  }
  if (num[x] == low[x]) {
    pair<int, int> cur(((1 << 29) + 123), -((1 << 29) + 123));
    while (stdfs.size() && num[stdfs.top()] >= num[x]) {
      cur.first = min(cur.first, stdfs.top());
      cur.second = max(cur.second, stdfs.top());
      id[stdfs.top()] = bcc;
      stdfs.pop();
    }
    if (cur.second != cur.first) st[cur.second] = cur.first;
    bcc++;
  }
}
int main() {
  if (fopen("data.in", "r")) freopen("data.in", "r", stdin);
  scanf("%d", &N);
  scanf("%d", &M);
  for (int i = 0; i < M; i++) {
    int a, b;
    scanf("%d", &a);
    scanf("%d", &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 1; i <= N; i++)
    if (!num[i]) dfs(i, i);
  scanf("%d", &Q);
  for (int i = 0; i < Q; i++) {
    scanf("%d %d", &q[i][0], &q[i][1]);
    pq1.push(make_pair(-q[i][0], i));
  }
  for (int l = 1, r = 1; l <= N; l++) {
    while (pq1.size() && -pq1.top().first == l) {
      int id = pq1.top().second;
      pq1.pop();
      pq2.push(make_pair(-q[id][1], id));
      ans[id] = -tree.query(1, 1, N, q[id][0], q[id][1]);
    }
    while (r <= N && st[r] < l) r++;
    tree.update(1, 1, N, l, r - 1, 1);
    while (pq2.size() && -pq2.top().first == l) {
      int id = pq2.top().second;
      pq2.pop();
      ans[id] += tree.query(1, 1, N, q[id][0], q[id][1]);
    }
  }
  for (int i = 0; i < Q; i++) {
    printf("%I64d\n", ans[i]);
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int N, M, id[300011], bcc = 1;
int num[300011], low[300011], st[300011];
int Q, q[300011][2];
long long csum, inc1, inc2, change[300011], ans[300011];
priority_queue<pair<int, int> > pq1;
priority_queue<pair<int, int> > pq2;
vector<int> adj[300011];
stack<int> stdfs;
priority_queue<pair<int, int> > rnge;
struct segtree {
  long long data[4 * 300011], lazy[4 * 300011];
  void push(int w, int L, int R) {
    if (!lazy[w])
      return;
    else if (L != R) {
      lazy[w * 2] += lazy[w];
      lazy[w * 2 + 1] += lazy[w];
    }
    data[w] += lazy[w] * 1LL * (R - L + 1);
    lazy[w] = 0;
  }
  void update(int w, int L, int R, int a, int b, int v) {
    push(w, L, R);
    if (b < L || a > R)
      return;
    else if (a <= L && b >= R) {
      lazy[w] += v;
      push(w, L, R);
    } else {
      update(w * 2, L, (L + R) / 2, a, b, v);
      update(w * 2 + 1, (L + R) / 2 + 1, R, a, b, v);
      data[w] = data[w * 2] + data[w * 2 + 1];
    }
  }
  long long query(int w, int L, int R, int a, int b) {
    push(w, L, R);
    if (b < L || a > R)
      return 0;
    else if (a <= L && b >= R)
      return data[w];
    else {
      long long q1 = query(w * 2, L, (L + R) / 2, a, b);
      long long q2 = query(w * 2 + 1, (L + R) / 2 + 1, R, a, b);
      return q1 + q2;
    }
  }
} tree;
void dfs(int x, int p) {
  static int cnt = 0;
  bool isAp = false;
  stdfs.push(x);
  num[x] = low[x] = (++cnt);
  for (int i = 0; i < (int)adj[x].size(); i++) {
    int nxt = adj[x][i];
    if (nxt == p) continue;
    if (!num[nxt]) {
      dfs(nxt, x);
      isAp |= (low[nxt] >= num[x]);
    }
    low[x] = min(low[x], low[nxt]);
  }
  if (num[x] == low[x]) {
    pair<int, int> cur(((1 << 29) + 123), -((1 << 29) + 123));
    while (stdfs.size() && num[stdfs.top()] >= num[x]) {
      cur.first = min(cur.first, stdfs.top());
      cur.second = max(cur.second, stdfs.top());
      id[stdfs.top()] = bcc;
      stdfs.pop();
    }
    if (cur.second != cur.first) st[cur.second] = cur.first;
    bcc++;
  }
}
int main() {
  if (fopen("data.in", "r")) freopen("data.in", "r", stdin);
  scanf("%d", &N);
  scanf("%d", &M);
  for (int i = 0; i < M; i++) {
    int a, b;
    scanf("%d", &a);
    scanf("%d", &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 1; i <= N; i++)
    if (!num[i]) dfs(i, i);
  scanf("%d", &Q);
  for (int i = 0; i < Q; i++) {
    scanf("%d %d", &q[i][0], &q[i][1]);
    pq1.push(make_pair(-q[i][0], i));
  }
  for (int l = 1, r = 1; l <= N; l++) {
    while (pq1.size() && -pq1.top().first == l) {
      int id = pq1.top().second;
      pq1.pop();
      pq2.push(make_pair(-q[id][1], id));
      ans[id] = -tree.query(1, 1, N, q[id][0], q[id][1]);
    }
    while (r <= N && st[r] < l) r++;
    tree.update(1, 1, N, l, r - 1, 1);
    while (pq2.size() && -pq2.top().first == l) {
      int id = pq2.top().second;
      pq2.pop();
      ans[id] += tree.query(1, 1, N, q[id][0], q[id][1]);
    }
  }
  for (int i = 0; i < Q; i++) {
    printf("%I64d\n", ans[i]);
  }
  return 0;
}

```