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 INF = 0x3f3f3f3f;
const long long mod = (int)1e9 + 7;
const int N = 2e5 + 100;
int l[N], r[N], belong[N];
long long lz[N], sum[N], v[N], a[N], tmp[N];
int n, m, tot;
void Build() {
  m = sqrt(n);
  tot = n / m;
  if (n % m) tot++;
  for (int i = 1; i <= n; i++) belong[i] = (i - 1) / m + 1, a[i] = i;
  for (int i = 1; i <= tot; i++) l[i] = (i - 1) * m + 1, r[i] = i * m;
  r[tot] = n;
}
void PushDown(int rt) {
  for (int i = l[rt]; i <= r[rt]; i++) a[i] = lz[rt];
  lz[rt] = 0;
}
void Update(int x, int y, long long c) {
  int idx = belong[x], idy = belong[y];
  if (idx == idy) {
    if (lz[idx]) PushDown(idx);
    for (int i = x; i <= y; i++) {
      sum[idx] += abs(c - a[i]);
      v[i] += abs(c - a[i]);
      a[i] = c;
    }
  } else {
    if (lz[idx]) PushDown(idx);
    if (lz[idy]) PushDown(idy);
    for (int i = x; i <= r[idx]; i++) {
      sum[idx] += abs(c - a[i]);
      v[i] += abs(c - a[i]);
      a[i] = c;
    }
    for (int i = l[idy]; i <= y; i++) {
      sum[idy] += abs(c - a[i]);
      v[i] += abs(c - a[i]);
      a[i] = c;
    }
    for (int i = idx + 1; i < idy; i++) {
      if (lz[i]) {
        tmp[i] += abs(lz[i] - c);
        sum[i] += (r[i] - l[i] + 1) * (abs(lz[i] - c));
        lz[i] = c;
      } else {
        lz[i] = c;
        for (int j = l[i]; j <= r[i]; j++) {
          v[j] += abs(a[j] - c);
          sum[i] += abs(a[j] - c);
          a[j] = c;
        }
      }
    }
  }
}
void Query(int x, int y) {
  int idx = belong[x], idy = belong[y];
  long long ans = 0;
  if (idx == idy) {
    for (int i = x; i <= y; i++) ans += (tmp[idx] + v[i]);
  } else {
    for (int i = x; i <= r[idx]; i++) ans += (tmp[idx] + v[i]);
    for (int i = l[idy]; i <= y; i++) ans += (tmp[idy] + v[i]);
    for (int i = idx + 1; i < idy; i++) ans += sum[i];
  }
  printf("%lld\n", ans);
}
int main() {
  int q, x, y, c, op;
  scanf("%d%d", &n, &q);
  Build();
  while (q--) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      scanf("%d", &c);
      Update(x, y, c);
    } else {
      Query(x, y);
    }
  }
  return 0;
}
