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;
#pragma comment(linker, "/STACK:102400000,102400000")
const int maxn = 1e5 + 5;
long long a[maxn], b[maxn], add[maxn], sum[maxn], lazy[maxn], lastval[maxn];
int block, num, belong[maxn], l[maxn], r[maxn];
int n, m;
void build() {
  block = sqrt(n);
  num = n / block;
  if (n % block) num++;
  for (int i = 1; i <= num; i++) l[i] = (i - 1) * block + 1, r[i] = i * block;
  for (int i = 1; i <= n; i++) belong[i] = (i - 1) / block + 1;
}
void update(int x, int y, int val) {
  if (belong[x] == belong[y]) {
    if (lastval[belong[x]]) {
      for (int i = l[belong[x]]; i <= r[belong[x]]; i++)
        a[i] = lastval[belong[x]];
      lastval[belong[x]] = 0;
    }
    for (int i = x; i <= y; i++) {
      b[i] += abs(val - a[i]);
      sum[belong[x]] += abs(val - a[i]);
      a[i] = val;
    }
    return;
  }
  if (lastval[belong[x]]) {
    for (int i = l[belong[x]]; i <= r[belong[x]]; i++)
      a[i] = lastval[belong[x]];
    lastval[belong[x]] = 0;
  }
  for (int i = x; i <= r[belong[x]]; i++) {
    b[i] += abs(val - a[i]);
    sum[belong[x]] += abs(val - a[i]);
    a[i] = val;
  }
  if (lastval[belong[y]]) {
    for (int i = l[belong[y]]; i <= r[belong[y]]; i++)
      a[i] = lastval[belong[y]];
    lastval[belong[y]] = 0;
  }
  for (int i = l[belong[y]]; i <= y; i++) {
    b[i] += abs(val - a[i]);
    sum[belong[y]] += abs(val - a[i]);
    a[i] = val;
  }
  for (int i = belong[x] + 1; i < belong[y]; i++) {
    if (lastval[i]) {
      lazy[i] += abs(val - lastval[i]);
      sum[i] += 1LL * (r[i] - l[i] + 1) * abs(val - lastval[i]);
      lastval[i] = val;
    } else {
      for (int j = l[i]; j <= r[i]; j++) {
        b[j] += abs(val - a[j]);
        sum[i] += abs(val - a[j]);
        a[j] = val;
      }
      lastval[i] = val;
    }
  }
}
long long query(int x, int y) {
  long long ans = 0;
  if (belong[x] == belong[y]) {
    for (int i = x; i <= y; i++) ans += b[i] + lazy[belong[x]];
    return ans;
  }
  for (int i = x; i <= r[belong[x]]; i++) ans += b[i] + lazy[belong[x]];
  for (int i = l[belong[y]]; i <= y; i++) ans += b[i] + lazy[belong[y]];
  for (int i = belong[x] + 1; i < belong[y]; i++) ans += sum[i];
  return ans;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) a[i] = i;
  build();
  for (int i = 1; i <= m; i++) {
    int ty, x, y, z;
    scanf("%d", &ty);
    if (ty == 1) {
      scanf("%d%d%d", &x, &y, &z);
      update(x, y, z);
    }
    if (ty == 2) {
      scanf("%d%d", &x, &y);
      printf("%lld\n", query(x, y));
    }
  }
  return 0;
}
