description: CQXYM wants to create a connected undirected graph with n nodes and m edges, and the diameter of the graph must be strictly less than k-1. Also, CQXYM doesn't want a graph that contains self-loops or multiple edges (i.e. each edge connects two different vertices and between each pair of vertices there is at most one edge).
  
  The diameter of a graph is the maximum distance between any two nodes.
  
  The distance between two nodes is the minimum number of the edges on the path which endpoints are the two nodes.
  
  CQXYM wonders whether it is possible to create such a graph.
  
  Input
  
  The input consists of multiple test cases. 
  
  The first line contains an integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of the test cases follows.
  
  Only one line of each test case contains three integers n(1 ≤ n ≤ 10^9), m, k (0 ≤ m,k ≤ 10^9).
  
  Output
  
  For each test case, print YES if it is possible to create the graph, or print NO if it is impossible. You can print each letter in any case (upper or lower).
  
  Example
  
  Input
  
  
  5
  1 0 3
  4 5 3
  4 6 3
  5 4 1
  2 1 1
  
  
  Output
  
  
  YES
  NO
  YES
  NO
  NO
  
  Note
  
  In the first test case, the graph's diameter equal to 0.
  
  In the second test case, the graph's diameter can only be 2.
  
  In the third test case, the graph's diameter can only be 1.
solution: import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.zip.CheckedInputStream;
  
  import javax.naming.InitialContext;
  
  import java.util.*;
  
  public class Main {
  
  	static long mod = 1_000_000_007;
  
  	public static void main(String[] args) {
  
  		FastScanner f = new FastScanner();
  		PrintWriter p = new PrintWriter(System.out);
  		int t=f.nextInt();
  		while(t-->0) {
  			long v=f.nextLong();
  			long e=f.nextLong();
  			long d=f.nextLong();
  			boolean res=solver2(v, e, d);
  			p.println(res==true?"YES":"NO");
  		}
  		p.close();
  
  
  	}
  	
  	static boolean solver(int v,int e,int d) {
  		
  		//if graph contains a single node then diameter is 0
  		if(v==1) {
  			if(d>1) return true;
  			else return false;
  		}
  		
  		
  		//if number of edges are less than vertices then graph cannot be connected
  		if(e<v-1&&v>1) return false;
  		
  		//if graph is a complete graph but contains multiedges i.e e>n*(n-1)/2
  		if(e>(v*(v-1)/2)&&v>1) return false;
  		
  		
  
  		
  		//if graph is a complete graph then diameter=1; number of edges=n*(n-1)/2
  		if(e==(v*(v-1)/2)&&v>1) {
  			if(d>2) return true;
  			return false;
  		}
  		
  		
  		//if the graph is a tree then diameter is atleast 2
  		if(e==v-1&&e<(v*(v-1)/2)&&v>1) {
  			if(d>3) return true;
  			return false;
  		}
  		
  		return false;
  		
  		
  	}
  	
  	static boolean solver2(long v,long e,long d) {
  		
  		if((v-1)*v>>1<e||e<v-1){
  			return false;
  		}
  		if(v==1){
  			if(d>1){
  				return true;
  			}else{
  				return false;
  			}
  		}else if(e<(v-1)*v>>1){
  			if(d>3){
  				return  true;
  			}else{
  				return false;
  			}
  		}else if(d>2){
  			return true;
  		}else{
  			return false;
  		}
  	}
  	
  	
  	
  	
  	
  	
  	
  	
  	
  
  	
  
  //-----------------------------------------------------------------------------------------------------------------------------------------------------------------
  
  	static int offx[] = { -1, -1, -1, 0, 1, 1, 1, 0 };
  	static int offy[] = { -1, 0, 1, 1, 1, 0, -1, -1 };
  
  	static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastScanner() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		String next() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		String nextLine() {
  			String str = "";
  			try {
  				str = br.readLine();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  	}
  
  	static long add(long x, long y) {
  		x += y;
  		if (x >= mod)
  			return (x % mod);
  		return x;
  	}
  
  	static long sub(long x, long y) {
  		x -= y;
  		if (x < 0)
  			return (x + mod);
  		return x;
  	}
  
  	static long mul(long x, long y) {
  		return (x * y) % mod;
  	}
  
  	static long bin_pow(long x, long p) {
  		if (p == 0)
  			return 1;
  		if ((p & 1) != 0)
  			return mul(x, bin_pow(x, p - 1));
  		return bin_pow(mul(x, x), p / 2);
  	}
  
  	static long rev(long x) {
  		return bin_pow(x, mod - 2);
  	}
  
  	static long div(long x, long y) {
  		return mul(x, rev(y));
  	}
  
  	// will work till array size of 65537---fastest sorting time
  //	static long[] radixSort(long[] f) {
  //		return radixSort(f, f.length);
  //	}
  //
  //	static long[] radixSort(long[] f, int n) {
  //		long[] to = new long[n];
  //		{
  //			int[] b = new int[65537];
  //			for (int i = 0; i < n; i++)
  //				b[1 + (int) (f[i] & 0xffff)]++;
  //			for (int i = 1; i <= 65536; i++)
  //				b[i] += b[i - 1];
  //			for (int i = 0; i < n; i++)
  //				to[b[(int) (f[i] & 0xffff)]++] = f[i];
  //			long[] d = f;
  //			f = to;
  //			to = d;
  //		}
  //		{
  //			int[] b = new int[65537];
  //			for (int i = 0; i < n; i++)
  //				b[1 + (int) (f[i] >>> 16 & 0xffff)]++;
  //			for (int i = 1; i <= 65536; i++)
  //				b[i] += b[i - 1];
  //			for (int i = 0; i < n; i++)
  //				to[b[(int) (f[i] >>> 16 & 0xffff)]++] = f[i];
  //			long[] d = f;
  //			f = to;
  //			to = d;
  //		}
  //		{
  //			int[] b = new int[65537];
  //			for (int i = 0; i < n; i++)
  //				b[1 + (int) (f[i] >>> 32 & 0xffff)]++;
  //			for (int i = 1; i <= 65536; i++)
  //				b[i] += b[i - 1];
  //			for (int i = 0; i < n; i++)
  //				to[b[(int) (f[i] >>> 32 & 0xffff)]++] = f[i];
  //			long[] d = f;
  //			f = to;
  //			to = d;
  //		}
  //		{
  //			int[] b = new int[65537];
  //			for (int i = 0; i < n; i++)
  //				b[1 + (int) (f[i] >>> 48 & 0xffff)]++;
  //			for (int i = 1; i <= 65536; i++)
  //				b[i] += b[i - 1];
  //			for (int i = 0; i < n; i++)
  //				to[b[(int) (f[i] >>> 48 & 0xffff)]++] = f[i];
  //			long[] d = f;
  //			f = to;
  //			to = d;
  //		}
  //		return f;
  //	}
  
  	// Function to sort an array using quick sort algorithm.
  	static void quickSort(long arr[], int low, int high) {
  		if (low < high) {
  			int p = partition(arr, low, high);
  			quickSort(arr, low, p - 1);
  			quickSort(arr, p + 1, high);
  		}
  	}
  
  	static int partition(long arr[], int low, int high) {
  		long pivot = arr[high];
  		int i = low - 1;
  		for (int j = low; j < high; j++) {
  			if (arr[j] < pivot) {
  				i++;
  				long temp = arr[i];
  				arr[i] = arr[j];
  				arr[j] = temp;
  			}
  		}
  		long temp = arr[i + 1];
  		arr[i + 1] = arr[high];
  		arr[high] = temp;
  
  		return i + 1;
  	}
  
  	/// Ceils
  	static long ceil(long x, long m) {
  		long res = x / m;
  		if (x % m != 0) {
  			res++;
  		}
  		return res;
  	}
  
  	// ------------------------------------------------------------------------------------------------
  	// makes the prefix sum array
  	static long[] prefixSum(long arr[], int n) {
  		long psum[] = new long[n];
  		psum[0] = arr[0];
  		for (int i = 1; i < n; i++) {
  			psum[i] = psum[i - 1] + arr[i];
  		}
  		return psum;
  	}
  	// ------------------------------------------------------------------------------------------------
  
  	// makes the suffix sum array
  	static long[] suffixSum(long arr[], int n) {
  		long ssum[] = new long[n];
  		ssum[n - 1] = arr[n - 1];
  		for (int i = n - 2; i >= 0; i--) {
  			ssum[i] = ssum[i + 1] + arr[i];
  		}
  		return ssum;
  	}
  
  //------------------------------------------------------------------------------------------
  	// BINARY EXPONENTIATION OF A NUMBER MODULO M FASTER METHOD WITHOUT RECURSIVE
  	// OVERHEADS
  //	static long m = (long) (1e9 + 7);
  
  	static long binPower(long a, long n, long m) {
  		if (n == 0)
  			return 1;
  		long res = 1;
  		while (n > 0) {
  			if ((n & 1) != 0) {
  				res *= a;
  			}
  			a *= a;
  			n >>= 1;
  		}
  		return res;
  	}
  
  //-------------------------------------------------------------------------------------------
  
  	// gcd
  	static long gcd(long a, long b) {
  		return b == 0 ? a : gcd(b, a % b);
  	}
  
  //------------------------------------------------------------------------------------------
  
  	// lcm
  	static long lcm(long a, long b) {
  		return a / gcd(a, b) * b;
  	}
  
  	// ------------------------------------------------------------------------------------------
  	// BRIAN KERNINGHAM TO CHECK NUMBER OF SET BITS
  	// O(LOGn)
  	static int setBits(int n) {
  		int count = 0;
  		while (n > 0) {
  			n = n & (n - 1);
  			count++;
  		}
  		return count;
  	}
  //------------------------------------------------------------------------------------------
  
  	// ------------------------------------------------------------------------------------------
  	// 0 based indexing
  	static boolean KthBitSet(int n, int k) {
  		int mask = 1;
  		mask = mask <<= k;
  		if ((mask & n) != 0)
  			return true;
  		else
  			return false;
  	}
  
  //------------------------------------------------------------------------------------------
  
  	// EXTENDED EUCLIDEAN THEOREM
  	// TO REPRESENT GCD IN TERMS OF A AND B
  	// gcd(a,b) = a.x + b.y where x and y are integers
  
  	static long x = -1;
  	static long y = -1;
  
  	static long gcdxy(long a, long b) {
  		if (b == 0) {
  			x = 1;
  			y = 0;
  			return a;
  		} else {
  			long d = gcdxy(b, a % b);
  			long x1 = y;
  			long y1 = x - (a / b) * y;
  			x = x1;
  			y = y1;
  
  			return d;
  		}
  	}
  
  //-------------------------------------------------------------------------------------------------F
  }
  