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 long long Maxn = 500005;
struct node {
  long long sum;
  long long colour;
  long long add;
} w[Maxn * 4];
long long n, m;
void Build(long long L, long long R, long long now) {
  w[now].sum = w[now].colour = w[now].add = 0;
  if (L == R) {
    w[now].colour = L;
    return;
  }
  long long mid = (L + R) >> 1;
  Build(L, mid, now * 2);
  Build(mid + 1, R, now * 2 + 1);
}
void Pushdown(long long L, long long R, long long now) {
  long long mid = (L + R) >> 1;
  w[now * 2].add += w[now].add;
  w[now * 2 + 1].add += w[now].add;
  w[now * 2].sum += w[now].add * (mid - L + 1);
  w[now * 2 + 1].sum += w[now].add * (R - mid);
  w[now].add = 0;
}
long long Query(long long l, long long r, long long L, long long R,
                long long now) {
  if (L > r || R < l) return 0;
  if (L >= l && R <= r) return w[now].sum;
  Pushdown(L, R, now);
  long long mid = (L + R) >> 1;
  return Query(l, r, L, mid, 2 * now) + Query(l, r, mid + 1, R, 2 * now + 1);
}
void Change(long long l, long long r, long long c, long long L, long long R,
            long long now) {
  if (L > r || R < l) return;
  if (l <= L && r >= R && w[now].colour) {
    long long len = R - L + 1;
    w[now].add += abs(w[now].colour - c);
    w[now].sum += len * abs(w[now].colour - c);
    w[now].colour = c;
    return;
  }
  Pushdown(L, R, now);
  if (w[now].colour) w[now * 2].colour = w[now * 2 + 1].colour = w[now].colour;
  long long mid = (L + R) >> 1;
  Change(l, r, c, L, mid, now * 2);
  Change(l, r, c, mid + 1, R, now * 2 + 1);
  w[now].colour =
      w[2 * now].colour == w[2 * now + 1].colour ? w[2 * now].colour : 0;
  w[now].sum = w[2 * now].sum + w[2 * now + 1].sum;
}
void Solve() {
  scanf("%I64d%I64d", &n, &m);
  Build(1, n, 1);
  for (long long i = 1; i <= m; i++) {
    long long f, l, r, c;
    scanf("%I64d%I64d%I64d", &f, &l, &r);
    if (f == 1) {
      scanf("%I64d", &c);
      Change(l, r, c, 1, n, 1);
    } else {
      c = Query(l, r, 1, n, 1);
      printf("%I64d\n", c);
    }
  }
}
int main() {
  Solve();
  return 0;
}
