Scala Spark skeleton implementing grep

pom.xml

<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
	<modelVersion>4.0.0</modelVersion>
	<groupId>my.org</groupId>
	<artifactId>spark-grep</artifactId>
	<version>1.0-SNAPSHOT</version>
	<dependencies>
		<dependency>
			<groupId>org.apache.spark</groupId>
			<artifactId>spark-core_2.10</artifactId>
			<version>1.2.0</version>
		</dependency>
	</dependencies>
	<build>
		<plugins>
			<plugin>
				<groupId>org.apache.maven.plugins</groupId>
				<artifactId>maven-compiler-plugin</artifactId>
				<version>3.1</version>
			</plugin>
		</plugins>
	</build>
</project>

SparkGrep.scala

package spark.example

import org.apache.spark.SparkContext
import org.apache.spark.SparkContext._
import org.apache.spark.SparkConf

object SparkGrep {
	def main(args: Array[String]) {
		if (args.length < 3) {
			System.err.println("Usage: SparkGrep <host> <input_file> <match_term>")
			System.exit(1)
		}
		val conf = new SparkConf().setAppName("SparkGrep").setMaster(args(0))
		val sc = new SparkContext(conf)
		val inputFile = sc.textFile(args(1), 2).cache()
		val matchTerm : String = args(2)
		val numMatches = inputFile.filter(line => line.contains(matchTerm)).count()
		println("%s lines in %s contain %s".format(numMatches, args(1), matchTerm))
		System.exit(0)
	}
}

I ran this code on 700 files giving argument(2) as //.txt . I want the output as 4 lines in //part-123.txt contain . but instead of that I am getting 4 lines in //.txt contain how I will get to know the File name out of this 700 files where my search variable is . Can you please help me out here. Will really be helpful from your end.