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;
long long color[100005 << 2];
long long sum[100005 << 2];
long long delta[100005 << 2];
int n, m, ql, qr, x;
long long ans;
void build(int o, int L, int R) {
  if (L == R) {
    sum[o] = delta[o] = 0;
    color[o] = L;
    return;
  }
  color[o] = 0;
  int mid = (L + R) >> 1;
  build(o << 1, L, mid);
  build(o << 1 | 1, mid + 1, R);
}
void pushup(int o, int L, int R) {
  if (color[o << 1] == color[o << 1 | 1])
    color[o] = color[o << 1];
  else
    color[o] = 0;
  sum[o] = sum[o << 1] + sum[o << 1 | 1];
}
void pushdown(int o, int L, int R) {
  if (!color[o]) return;
  int mid = (R + L) >> 1;
  delta[o << 1] += delta[o];
  delta[o << 1 | 1] += delta[o];
  sum[o << 1] += delta[o] * (mid - L + 1);
  sum[o << 1 | 1] += delta[o] * (R - mid);
  color[o << 1] = color[o << 1 | 1] = color[o];
  delta[o] = color[o] = 0;
}
void updata(int o, int L, int R) {
  if (ql <= L && qr >= R && color[o]) {
    delta[o] += abs(color[o] - x);
    sum[o] += abs(color[o] - x) * (R - L + 1);
    color[o] = x;
    return;
  }
  pushdown(o, L, R);
  int mid = (L + R) >> 1;
  if (ql <= mid) updata(o << 1, L, mid);
  if (qr > mid) updata(o << 1 | 1, mid + 1, R);
  pushup(o, L, R);
}
void query(int o, int L, int R) {
  if (ql <= L && qr >= R) {
    ans += sum[o];
    return;
  }
  pushdown(o, L, R);
  int mid = (L + R) >> 1;
  if (ql <= mid) query(o << 1, L, mid);
  if (qr > mid) query(o << 1 | 1, mid + 1, R);
  pushup(o, L, R);
}
void debug(void) { printf("%I64d %I64d\n", sum[4], sum[5]); }
void solve(void) {
  int k;
  while (m--) {
    scanf("%d", &k);
    if (k == 1) {
      scanf("%d%d%d", &ql, &qr, &x);
      updata(1, 1, n);
    } else {
      scanf("%d%d", &ql, &qr);
      ans = 0;
      query(1, 1, n);
      printf("%I64d\n", ans);
    }
  }
}
int main(void) {
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  solve();
  return 0;
}
