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 MAXN = 100000 + 10;
int belong[MAXN], block, num, L[MAXN], R[MAXN];
int n, q;
long long int a[MAXN], b[MAXN];
struct Node {
  long long int lazy, Bsum, lazyb;
} Bval[400];
void build() {
  block = (int)sqrt(n + 0.5);
  num = n / block;
  if (n % block) {
    num++;
  }
  for (int i = 1; i <= num; i++) {
    Bval[i].Bsum = 0;
    Bval[i].lazy = 0;
    Bval[i].lazyb = 0;
    L[i] = (i - 1) * block + 1;
    R[i] = i * block;
  }
  R[num] = n;
  for (int i = 1; i <= n; i++) {
    belong[i] = ((i - 1) / block) + 1;
  }
}
void modify(int st, int ed, int val) {
  if (belong[st] == belong[ed]) {
    if (Bval[belong[st]].lazy) {
      for (int i = L[belong[st]]; i <= R[belong[st]]; i++) {
        a[i] = Bval[belong[st]].lazy;
      }
      Bval[belong[st]].lazy = 0;
    }
    for (int i = st; i <= ed; i++) {
      Bval[belong[st]].Bsum += abs(a[i] - val);
      b[i] += abs(a[i] - val);
      a[i] = val;
    }
    return;
  }
  if (Bval[belong[st]].lazy) {
    for (int i = L[belong[st]]; i <= R[belong[st]]; i++) {
      a[i] = Bval[belong[st]].lazy;
    }
    Bval[belong[st]].lazy = 0;
  }
  for (int i = st; i <= R[belong[st]]; i++) {
    Bval[belong[st]].Bsum += abs(val - a[i]);
    b[i] += abs(val - a[i]);
    a[i] = val;
  }
  for (int i = belong[st] + 1; i < belong[ed]; i++) {
    if (Bval[i].lazy) {
      Bval[i].lazyb += abs(val - Bval[i].lazy);
      Bval[i].Bsum += (1LL * abs(Bval[i].lazy - val) * (R[i] - L[i] + 1));
      Bval[i].lazy = val;
    } else {
      for (int j = L[i]; j <= R[i]; j++) {
        b[j] += abs(a[j] - val);
        Bval[i].Bsum += abs(a[j] - val);
        a[j] = val;
      }
      Bval[i].lazy = val;
    }
  }
  if (Bval[belong[ed]].lazy) {
    for (int i = L[belong[ed]]; i <= R[belong[ed]]; i++) {
      a[i] = Bval[belong[ed]].lazy;
    }
    Bval[belong[ed]].lazy = 0;
  }
  for (int i = L[belong[ed]]; i <= ed; i++) {
    Bval[belong[ed]].Bsum += abs(val - a[i]);
    b[i] += abs(val - a[i]);
    a[i] = val;
  }
}
long long int query(int st, int ed) {
  long long int ans = 0;
  if (belong[st] == belong[ed]) {
    for (int i = st; i <= ed; i++) {
      ans += (b[i] + Bval[belong[st]].lazyb);
    }
    return ans;
  }
  for (int i = st; i <= R[belong[st]]; i++) {
    ans += (b[i] + Bval[belong[st]].lazyb);
  }
  for (int i = belong[st] + 1; i < belong[ed]; i++) {
    ans += Bval[i].Bsum;
  }
  for (int i = L[belong[ed]]; i <= ed; i++) {
    ans += (b[i] + Bval[belong[ed]].lazyb);
  }
  return ans;
}
int main() {
  while (~scanf("%d%d", &n, &q)) {
    for (int i = 1; i <= n; i++) {
      a[i] = i;
      b[i] = 0;
    }
    build();
    int type, l, r, v;
    for (int i = 1; i <= q; i++) {
      scanf("%d", &type);
      if (type == 1) {
        scanf("%d%d%d", &l, &r, &v);
        modify(l, r, v);
      } else {
        scanf("%d%d", &l, &r);
        printf("%lld\n", query(l, r));
      }
    }
  }
  return 0;
}
