DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

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


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
long long RT[N << 2], LZNUM[N << 2], LZ[N << 2], MX[N << 2], MN[N << 2];
int n, q, tp;
void Do(int id) {
  MN[id] = min(MN[(id << 1)], MN[(id << 1 | 1)]);
  MX[id] = max(MX[(id << 1)], MX[(id << 1 | 1)]);
  RT[id] = RT[(id << 1)] + RT[(id << 1 | 1)];
}
void build(int id = 1, int l = 0, int r = n) {
  if (r - l < 2) {
    MX[id] = MN[id] = r;
    return;
  }
  int mid = ((l + r) >> 1);
  build((id << 1), l, mid);
  build((id << 1 | 1), mid, r);
  Do(id);
}
void shift(int id, int l, int r) {
  if (r - l < 2 || !LZ[id]) return;
  int mid = ((l + r) >> 1);
  LZ[(id << 1)] = LZ[(id << 1 | 1)] = MX[(id << 1)] = MN[(id << 1)] =
      MX[(id << 1 | 1)] = MN[(id << 1 | 1)] = LZ[id];
  RT[(id << 1)] += 1LL * LZNUM[id] * (mid - l);
  RT[(id << 1 | 1)] += 1LL * LZNUM[id] * (r - mid);
  LZNUM[(id << 1)] += LZNUM[id];
  LZNUM[(id << 1 | 1)] += LZNUM[id];
  LZ[id] = LZNUM[id] = 0;
}
void add(int s, int e, int x, int id = 1, int l = 0, int r = n) {
  shift(id, l, r);
  if (l >= e || r <= s) return;
  if (l >= s && r <= e && MX[id] == MN[id]) {
    LZ[id] = x;
    LZNUM[id] += abs(MN[id] - x);
    RT[id] += 1LL * (r - l) * abs(MN[id] - x);
    MN[id] = MX[id] = x;
    return;
  }
  int mid = ((l + r) >> 1);
  add(s, e, x, (id << 1), l, mid);
  add(s, e, x, (id << 1 | 1), mid, r);
  Do(id);
}
long long get(int s, int e, int id = 1, int l = 0, int r = n) {
  if (l >= e || r <= s) return (0LL);
  if (l >= s && r <= e) return (RT[id]);
  shift(id, l, r);
  int mid = ((l + r) >> 1);
  return (get(s, e, (id << 1), l, mid) + get(s, e, (id << 1 | 1), mid, r));
}
int main() {
  scanf("%d %d", &n, &q);
  build();
  for (int Q = 0; Q < q; Q++) {
    scanf("%d", &tp);
    if (tp == 1) {
      int l, r, x;
      scanf("%d %d %d", &l, &r, &x);
      add(l - 1, r, x);
      continue;
    }
    int l, r;
    scanf("%d %d", &l, &r);
    printf("%lld\n", get(l - 1, r));
  }
}
