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 double EPS = 0.0000001;
const double PI = acos(-1);
const long long INFLL = 0x7FFFFFFFFFFFFFFF;
const int INF = 0x7FFFFFFF;
template <typename T>
inline void next(T &num) {
  char c;
  num = 0;
  do {
    c = getchar_unlocked();
  } while (c != EOF && c == ' ' && c == '\n' && c == '\t');
  int sign = (c == '-' ? -1 : 1);
  if (c != '-') num += (c - '0');
  while ((c = getchar_unlocked()) != EOF && c != '\n' && c != '\t' &&
         c != ' ') {
    num *= 10;
    num += (c - '0');
  }
  num *= sign;
}
inline string getstr() {
  string str;
  char k;
  while ((k = getchar_unlocked()) == ' ' || k == '\n') {
    k = getchar_unlocked();
    if (k == ' ' || k == '\n')
      continue;
    else
      break;
  }
  str.push_back(k);
  while ((k = getchar_unlocked()) != EOF && k != '\n' && k != '\t' &&
         k != '\v' && k != '\0' && k != ' ')
    str.push_back(k);
  return str;
}
const int N = 100000;
long long prop[4 * N], inc[4 * N];
int n, m, M = 1, Lb[4 * N], Rb[4 * N], color[4 * N], x;
void init(int l, int r, int cur) {
  Lb[cur] = l;
  Rb[cur] = r;
  if (l == r) {
    color[cur] = l;
    return;
  }
  int mid = (l + r) >> 1;
  init(l, mid, cur << 1);
  init(mid + 1, r, cur << 1 | 1);
}
void propagate(int l, int r, int cur) {
  if (color[cur]) {
    prop[cur] += abs(color[cur] - x);
    inc[cur] += (long long)(r - l + 1) * abs(color[cur] - x);
  } else {
    int mid = (l + r) >> 1;
    propagate(l, mid, cur << 1);
    propagate(mid + 1, r, cur << 1 | 1);
    inc[cur] = inc[cur << 1] + inc[cur << 1 | 1] + prop[cur] * (r - l + 1);
  }
}
void upd(int l, int r, int x, int cur) {
  if (l > Rb[cur] || r < Lb[cur]) return;
  if (l <= Lb[cur] && r >= Rb[cur]) {
    propagate(Lb[cur], Rb[cur], cur);
    color[cur] = x;
    return;
  }
  int mid = (Lb[cur] + Rb[cur]) >> 1;
  if (color[cur] > 0) {
    color[cur << 1] = color[cur];
    color[cur << 1 | 1] = color[cur];
    color[cur] = 0;
  }
  upd(l, min(mid, r), x, cur << 1);
  upd(max(l, mid + 1), r, x, cur << 1 | 1);
  color[cur] = 0;
  inc[cur] =
      inc[cur << 1] + inc[cur << 1 | 1] + prop[cur] * (Rb[cur] - Lb[cur] + 1);
}
long long get(int l, int r, int cur) {
  if (l > Rb[cur] || r < Lb[cur]) return 0;
  if (l <= Lb[cur] && r >= Rb[cur]) return inc[cur];
  int mid = (Lb[cur] + Rb[cur]) >> 1;
  long long ans = get(l, min(mid, r), cur << 1);
  ans += get(max(mid + 1, l), r, cur << 1 | 1);
  ans += prop[cur] * (r - l + 1);
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  while (M < n) M <<= 1;
  memset(color, 0, sizeof(color));
  init(1, M, 1);
  while (m--) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) {
      scanf("%d", &x);
      upd(l, r, x, 1);
    } else {
      printf("%I64d\n", get(l, r, 1));
    }
  }
  return 0;
}
